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RESISTANCE OF 
MATERIALS 



































































































































* 














































































RESISTANCE OF 
MATERIALS 


BY 

FRED B. SEELY, M.S. 

Professor of Theoretical and Applied Mechanics 
University of Illinois 


NEW YORK 

JOHN WILEY & SONS, Inc. 

London: CHAPMAN & HALL, Limited 
1924 



Copyright, 1924 

BY 

FRED B. SEELY 



Cl A 808496 


PRESS OF 

BRAUNWORTH & CO 
BOOK MANUFACTURERS 
BROOKLYN, N. Y. 


OCT 25 V 


PREFACE 


The subject of Resistance of Materials as developed in this 
book is divided in two parts. Part I, called Mechanics of Mate¬ 
rials, treats chiefly of the application of the principles of Analytical 
Mechanics and of the experimental laws of structural materials 
to the analysis of the action in the members used in structures and 
machines. Part II treats chiefly of the structural (force-resisting) 
properties of engineering materials. 

The main object of Part I is to develop rational methods for 
the design of the common types of force-resisting members used in 
engineering structures. The main objects of Part II are ( a) to 
investigate the properties of materials from which may be deter¬ 
mined the suitability of material for various structural uses, and 
( b ) to consider tests by means of which these properties may be 
measured. But, attention should be called to the fact that much 
of the value of the study of Resistance of Materials lies in gaining 
correct ideas or concepts of the ways various structural and 
machine members act in resisting loads, and of the adjustments 
that occur in a member as loads are applied to it; this general 
knowledge of the actions in materials and members has many 
opportunities to function in engineering practice where definite 
calculations are not required and when it is impossible to obtain 
quantitative results. 

Although Part I is self-contained, reference to Part II (as 
indicated throughout Part I) should prove of value to the student. 
Further, Part II should be of great help to the student in con¬ 
nection with laboratory work in materials testing. 

Throughout the book consideration is given to the effects, on 
the analysis, of deviations in the conditions that may obtain with 
actual members from those that are assumed in the analysis. 
Further, the limitations of the methods of analysis and of the 
formulas developed are pointed out. 



VI 


PREFACE 


Great care has been exercised in selecting problems that are of 
practical value and yet are easily comprehended and are free from 
unimportant details so that the principles used in their solution 
will stand out clearly. Illustrative problems are given at the 
end of the more important articles and many problems are offered 
for solution; the answers to about one-half of the problems are 
given. 

The moment-area principle for expressing the relations between 
the elastic properties of a beam and the external forces acting on 
the beam is treated in Chapters VIII and IX. Two methods of ap¬ 
plying or interpreting the moment-area principle are used; namely, 
the slope-deviation method and the conjugate-beam method. 
These methods are considered mainly to be supplementary to 
the double-integration method treated in Chapters VI and VII. 
Chapters VIII and IX may be omitted without destroying the 
continuity of the book, and either one of the two methods of 
interpretation may be studied without studying the other. 

Among the special features of the book may be mentioned the 
treatment of repeated stress and fatigue of metals (Chapter XIV); 
the emphasis on the principle of work and energy in determining 
the effect of impact loading (Chapter XIII); the use of the con¬ 
jugate-beam method, in addition to the slope-deviation method, 
in applying the moment-area principle to beams; and the dis¬ 
cussion, in Part II, of the significance of the structural properties 
of materials. 

During the preparation of the manuscript the author was 
greatly aided by valuable suggestions and contributions from many 
of his colleagues. The author is especially indebted to Professor 
A. N. Talbot for helpful suggestions in connection with the gen¬ 
eral point of view toward the subject, and also with the content 
of the book and the method of treatment of various topics; to 
Professor H. F. Moore for valuable contributions to the content 
of the Chapters on repeated stress and properties of materials; 
to Professor N. E. Ensign for reading the manuscript and for 
SDlving most of the problems; and to Professors H. M. Westergaard 
and F. E. Richart for contributions to the methods of applying the 
moment-area principle to beams. The help thus received has 
contributed much to the value of the book. 


August, 1924. 


F. B. S. 


CONTENTS 


PART I. MECHANICS OF MATERIALS 

Chapter I. Stress and Strain 

page 

1. Introduction. 1 

2. Types of Loading. 3 

Static Central Loads 

3. Stresses due to Central Loads. 5 

4. Strains due to Central Loads.. 10 

5. Stress-strain Curve. Proportional Limit. Yield-Point. Ultimate 

Strength. Elastic Limit. Modulus of Elasticity. 12 

6. Properties of Structural Materials. 18 

7. Working Stress. 23 

Chapter II. Thin-walled Cylinders and Spheres. Riveted 

Joints 

8. Stresses in Thin-walled Cylinders and Spheres. 30 

Riveted Joints 

9. Introduction. 34 

10. Types of Riveted Joints. Definitions.. 34 

11. Modes of Failure. 36 

12. Stresses in Riveted Joints. 37 

13. Assumptions. Conditions Affecting Strength of Riveted Joints. 43 

14. Allowable Stress. 46 

Chapter III. Elementary Combined Stresses and Combined 

Strains. Resilience 

15. Introduction. 48 

16. Stresses on Oblique Section. 48 

17. Shearing Stresses on Planes at Right Angles. Simple Shear. 50 

18. Tensile and Compressive Stresses Resulting from Simple Shear. 51 

vii 





















CONTENTS 


viii 

PAGE 

19. Principal Stresses. 53 

20. Shearing Stresses Resulting from Principal Stresses. 53 

21. Poisson’s Ratio. 55 

22. Strains Due to Principal Stresses. 55 

23. Resilience. 56 

Chapter IV. Stress and Deformation Due to Torsional Loads 

24. Torsional Load, Twisting Moment, and Resisting Moment Defined. . 60 

25. Distribution of Stress on the Cross-section. 61 

26. Expression for Resisting Moment. The Torsion Formula. 64 

27. Twisting Moment in Terms of Horsepower and Speed. 69 

28. Angle of Twist of Cylindrical Shaft. 70 

29. Shaft Couplings. 71 

30. Stress Beyond the Proportional Limit. Modulus of Rupture. 73 

31. Torsion of Non-circular Sections. 75 

Chapter V. Transverse Loads. Stresses in Statically Deter¬ 
minate Beams 

32. Preliminary Considerations. 77 

33. Vertical Shear, Resisting Shear, Bending Moment, and Resisting 

Moment. 80 

34. Expression for Resisting Moment. The Flexure Formula. 83 

35. Section of Maximum Bending Moment. 90 

36. Shear and Moment Diagrams. 92 

37. Relation Between Shear and Moment. 101 

38. Overhanging Beams. 102 

39. Economical Sections of Beams. 105 

40. Shearing Stress in a Beam. 107 

41. Stress Beyond Proportional Limit. Modulus of Rupture. 114 

42. Maximum Moment Due to Moving Loads. 116 

43. Assumptions and Limitations Involved in the Flexure Formula. 118 

Chapter VI. Deflection of Statically Determinate Beams 
(Double Integration Method) 

44. Introduction. 120 

45. Elastic Curve Equation. 120 

46. Simple Beam; Uniform Load. 125 

47. Simple Beam; Concentrated Load at Mid-span. 127 

48. Cantilever Beam; Uniform Load. 129 

49. Overhanging Beam; Uniform Load. 130 

50. Simple Beam; Concentrated Load not at Mid-span. 131 

































CONTENTS 


IX 


Chapter VII. Statically Indeterminate Beams 
Double Integration Method 

page 

51. Introduction. 135 

52. Beam Fixed at One End, Supported at Other End; Uniform Load. .. 136 

53. Beam Fixed at Both Ends; Uniform Load. 140 

54. Beam Fixed at Both Ends; Concentrated Load at Mid-Span. 143 

55. Beam Fixed at One End, Supported at Other End; Concentrated 

Load at Mid-span. 145 

56. Comparison of Fixed-ended and Simple Beams. 146 

57. Continuous Beams. Theorem of Three Moments. 148 

58. Solution of Typical Problem. 150 

59. Values of Moments and Shears in Continuous Beams. 153 

60. Advantages and Disadvantages of Continuous Beams. 155 

61. Theorem of Three Moments for Concentrated Loads. 155 

Chapter VIII. Deflection of Statically Determinate Beams 
(Moment-area Method) 

62. Introduction. 157 

Slope-deviation Method 

63. Theorems of the Slope-deviation Method. 158 

64. Simple Beam; Concentrated Load at Mid-span. 160 

65. Cantilever Beam; Concentrated Load at End. 161 

66. Simple Beam; Load Distributed Uniformly. 162 

67. Cantilever Beam; Load Distributed Uniformly. 163 

Conjugate-beam Method 

68. Conjugate Beam Defined. Equations Stated. 164 

69. Simple Beam; Concentrated Load at Mid-span. 166 

70. Cantilever Beam; Concentrated Load at End. 168 

71. Simple Beam; Load Distributed Uniformly. 169 

72. Cantilever Beam; Load Distributed Uniformly. 170 

73. Simple Beam; Concentrated Load at any Point. 170 

74. Simple Beam; Cross-section not Constant. 171 

Chapter IX. Statically Indeterminate Beams. (Moment-area 

Method) 


75. Introduction 


173 


























X 


CONTENTS 


Slope-deviation Method 

PAGE 

76. Beam Fixed at Both Ends; Load Concentrated at Mid-span. 173 

77. Beam Fixed at Both Ends; Uniform Load. 175 

78. Beam Fixed at One End, Supported at Other End; Concentrated 

Load at Mid-span. 175 

79. Beam Fixed at One End, Supported at Other End; Uniform Load. .. 176 

80. Continuous Beam. Theorem of Three Moments. 176 

Conjugate-beam Method 

81. Beam Fixed at Both Ends; Uniform Load. 178 

82. Beam Fixed at Both Ends; Concentrated Load at Mid-span. 180 

83. Beam Fixed at One End; Supported at Other End; Uniform Load. . 181 

84. Continuous Beam. Theorem of Three Moments. 181 

Chapter X. Combined Axial and Bending Loads. Eccentric 

Loads 

85. Introduction. 184 

86. A Beam Subjected to an Axial End Load. 184 

87. Eccentric Longitudinal Load in Plane of Symmetry. 188 

88. Eccentric Load not in Plane of Symmetry. 194 

89. Eccentric Loads on Riveted Connections. 196 

90. Helical Spring. 199 

Chapter XI. Compression Members. Columns 

91. Introduction. 204 

92. Distinguishing Features of Column Action. 205 

93. Slenderness Ratio. 207 

94. Two Limiting Cases of Compression Members. 207 

95. Euler’s Column Formula. 208 

96. Graphical Representation of Formulas for Ideal Columns. 214 

97. Methods of Obtaining Formulas for Columns Having Intermediate 

Slenderness Ratios. 217 

98. Gordon-Rankine Formula. 218 

99. Straight-line Formula. 224 

100. Effect of End Conditions. 230 

101. Rankine’s Formula for Columns with Restrained Ends. 233 

102. Straight-line Formula for Fixed-ended Columns. 234 

103. Straight-line Formula for High Carbon and Alloy Steel Columns.. . 235 

104. Eccentrically Loaded Columns. 237 

105. Equivalent Eccentricity. Secant Formula. 239 

106. Built-up Steel Columns. 245 

107. Columns Subjected to Cross-bending. 247 































CONTENTS 


xi 


Chapter XII. Combined Normal and Shearing Stresses 

page 

108. Introduction. 249 

109. Maximum Normal and Shearing Stresses. 250 

110. Maximum Normal Stress Occurs on Plane of Zero Shearing Stress. 

Ellipse of Stress. 259 

111. Diagonal Tensile Stress in a Beam. 260 

112. Expression for Ee . 263 

113. Theories of Failure of Elastic Action and Their Application.264 

Chapter XIII.— Impact and Energy Loads 

114. Introduction. 272 

115. Calculation of Energy Delivered to Resisting Member. 273 

116. Stress in Bar Due to Axial Energy Load. 275 

117. Comparison of Effects of Static and Energy Loads. 277 

1 18. Special Cases of Axial Energy Loads . 280 

119. Working Stress and Working Value of Energy. 282 

120. Stresses in Beams Due to Energy Loads. 288 

121. Effect of Form on Energy Resistance of Beams. 291 

122. Special Cases of Energy Loads on Beams. 293 

123. Deflection Due to any Energy Load. 294 

Chapter XIV. Repeated Loads. Fatigue of Metals 

124. Introduction. 297 

125. Endurance Limit. 298 

126. Localized Stress and Fatigue Failure. 300 

127. Values of Endurance Limits with Completely Reversed Bending 

Stress. 304 

128. Relation of Endurance Limit with Direct Axial Stress to Endurance 

Limit with Bending Stress. 305 

129. Relation of Endurance Limits in Torsion and Bending. 307 

130. Localised Stress Due to External Discontinuities. 307 

131. Working Stress with Repeated Loads. 309 

132. Effect of Range of Stress. 316 

PART II. MECHANICAL PROPERTIES OF STRUC¬ 
TURAL MATERIALS 

133. Introduction.. 324 

Chapter XV. Definitions, Methods of Measuring, and Signifi¬ 
cance of Mechanical Properties 

134. Properties to be Considered. 325 

135. Meaning of Strength. 325 

136. Static Ultimate Strength. 327 































CONTENTS 


xii 

PAGE 

137. Static Elastic Strength. 328 

138. The Significance of the Ultimate and Elastic Strengths. 343 

139. Static Compressive Strength. 345 

140. Static Shearing Strength. 347 

Ductility 

141. Definition of Ductility. 351 

142. Measure of Ductility. 351 

143. Significance and Need of Ductility. 353 

Stiffness 

144. Definition and Measure. 355 

145. Use and Significance of Stiffness. 359 

Resilience. Elastic Energy Strength 

146. Definition, Measure and Significance of Resilience. 360 

Toughness, Ultimate Impact or Ultimate Energy Strength 

147. Definition and Measure of Toughness. 363 

148. Significance of Toughness. 364 

149. The Single-blow Notched-bar Impact Test. 365 

Hardness 

150. Definition and Measure of Hardness. 367 

151. The Brinell Ball Test. 368 

152. The Shore Scleroscope. 368 

153. Limitations of the Tests. 369 

154. Need and Significance of Hardness. 371 

Fatigue Strength 

155. Definition and Measure of Fatigue Strength. 372 

156. Methods of Determining the Endurance Limit. 373 

Chapter XVI. Tables of Values of Properties of Materials 

157. Use of Tables. 379 

APPENDIX I. FIRST MOMENTS AND CENTROIDS OF 

AREAS 

158. Definitions. 391 

159. Centroids Found by Integration. 392 

160. Centroids of Composite Areas. 396 


























CONTENTS xiii 

APPENDIX II. SECOND MOMENT OR MOMENT OF INERTIA 

OF AN AREA 

PAGE 

161. Moment of Inertia of an Area Defined. 399 

162. Polar Moment of Inertia. 400 

163. Radius of Gyration.401 

164. Parallel Axis Theorem. 401 

165. Moments of Inertia Found by Integration. 403 

166. Moments of Inertia of Composite Areas. 407 

167. Approximate Method. 411 

APPENDIX III. LOCALIZED STRESS, ITS OCCURRENCE, 
SIGNIFICANCE AND MEASUREMENT 

168. Limitations of the Ordinary Methods of Analysis in Mechanics of 

Materials. 413 

169. Definition and Illustrations of Localized Stress.413 

170. The Significance of Localized Stress. 414 

171. Methods of Determining the Magnitude of Localized Stress. 415 




















. 























































RESISTANCE OF MATERIALS 


PART I. MECHANICS OF MATERIALS 


CHAPTER I 

STRESS AND STRAIN 

1. Introduction.—Resistance of Materials is that branch of 
Mechanics which treats of the internal forces in a physical body 
and of the changes of shape and size of the body, particularly in 
their relation to the external forces that act on the body, and to 
the physical properties of the material of the body. The external 
forces that act on the body are called loads ; the internal forces, 
which resist the external forces, are called stresses, and the changes 
in the dimensions of the body are called deformations or strains. ' 

The total stress 1 on a section through a body is the total internal 
force acting on the section; the component of the internal force 
acting normal to the area is called normal stress ; and the com¬ 
ponent acting tangent to (or in) the area is called shearing stress. 
Further, a normal stress may be a tensile stress or a compressive 
stress according as the body is stretched or is shortened. Intensity 
of stress or unit-stress 1 is defined to be stress per unit of area. In 
general the intensity of stress varies from point to point over a 
section, its value at any point being considered to be the stress on 
an elementary or differential part of the area, including the point, 
divided by the elementary area; but when the stress is distributed 
uniformly on an area the intensity of stress at all points in the sec¬ 
tion is equal to the total stress on the area divided by the whole 

I In technical literature the term “stress” is sometimes used to denote 
what is here defined as intensity of stress or unit-stress, and the term 
“internal force” is used to denote what is here called total stress. 




2 


STRESS AND STRAIN 


area. A unit-stress may be expressed in various units; such as 
pounds per square inch (lb. per sq. in.), kilograms per square cen¬ 
timeter (kg. per sq. cm.), tons per square inch (ton per sq. in.), 
etc. 


Similarly, the total deformation or total strain in any direction 
is the total change in the dimension of the body in that direction, 
and the unit-deformation or unit-strain in any direction is the defor¬ 
mation or strain per unit of length in that direction. 

The Problem Defined. —A body when subjected to loads is 
stressed and deformed, and the values of the stresses and the 
deformations in the body are of great importance in many engi¬ 
neering problems. These stresses are found by use of the general 
principles of mechanics (mainly of statics) and of the experimental 
laws that have been found to govern the action of the material. 
The main objects, then, in the study of Resistance of Materials 
are 

(1) To determine the relation between the external 
forces (loads) acting on a body and the resulting internal 
forces (stresses) in the material so that stresses may be 
determined from known loads, or so that the loads that 
produce give stresses may be found, and 

(2) To determine the relation between loads acting 
on a body and the resulting strains produced in the body 
so that strains may be determined from known loads, or 
vice versa, and 

(3) To obtain a knowledge of the physical properties, 
such as, stiffness, strength, ductility, toughness, resilience, 
etc., of the various structural materials, since physical 
properties of the materials are involved in the relations 
under (1) and (2) and are of special importance in determin¬ 
ing the suitability of a material for resisting loads under 
varying conditions of loading. 

f That part of Resistance of Materials which considers chiefly 
the application of the principles of mechanics to materials for the 
purposes stated under .(1) and (2) above is frequently called 
“ Mechanics of Materials ” and is discussed mainly in Part I of 
this book, whereas Part II is devoted to a brief discussion of the 
properties of structural materials. 


TYPES OF LOADING 


3 


As suggested by the statements under (1) and (2) above, in 
some problems of design, the stress produced in the body by the 
load is the governing factor in the design, whereas in other prob¬ 
lems the strain produced is the more important factor. For exam¬ 
ple, the chain of a hoist fulfills its function of lifting loads regard¬ 
less of the amount of stretch of the chain that occurs, provided 
that the load does not produce too large a stress in the chain. 
That is, the stress developed in the chain by the maximum load 
to be applied is the governing factor in the design of the chain. 
Likewise, stress is generally the governing factor in the design of 
many of the parts of bridges, buildings, cranes, ships, etc. In 
the design of machine tools, however, such as planers, lathes, drill 
presses, grinding machines, etc., the deformations of the parts 
are frequently of prime importance since such machines will not 
produce work of sufficient accuracy if the deformation of the parts 
is too large. 

In obtaining the relations under (1) and (2) above, it will be 
found that the stresses and strains produced in a body by the 
loads depend on (a) the type of loading (whether static loads, 
impact loads, or repeated loads), ( b ) the dimensions of the body or 
the shape of area on which the stress occurs, and (c) the properties 
of the material of the body. An example of the influence of each 
of the above factors may be given as follows: (a) a load applied 
suddenly to a body produces more stress and deformation in the 
body than does the same load when applied slowly; ( b ) a rolled 
I-section or channel-section has its area distributed better for 
resisting bending than does a bar of equal area having a rectan¬ 
gular or circular cross-section; (c) the elongation of an oak stick 
caused by a given load is greater than that of a bar of steel of the 
same dimensions when subjected to the same pull, etc., etc. 

2. Types of Loading.—With reference to the manner in which 
loads are applied or transmitted to a structure or machine, the 
loads will be considered under three distinct headings; namely, 
static loads, impact and energy loads, and repeated loads. 

1. Static , steady or dead loads are forces that are applied slowly 
and not repeated, and remain nearly constant after being applied 
to the body, or are repeated relatively few times; such as the 
loads on most buildings (a part of the load being the weights of the 
members of the structure), the load applied to a bar in a testing 
machine, etc. 


4 


STRESS AND STRAIN 


2. Impact loads are forces that are applied to the resisting body 
in a relatively short period of time; the shorter the time the 
greater the effect of the impact. An impact load, in general, is 
applied by a body that is in motion when it comes in contact with 
the resisting body, and the force exerted by the moving body and 
the period during which it acts can not in general be determined. 
For this reason in many problems it is more satisfactory to calcu¬ 
late the stress and strain produced by an impact load from the 
energy delivered to the resisting body by the moving body. When 
this is done the energy delivered to the resisting body is called an 
energy load and is expressed in foot-pounds (not in pounds). 
Impact and energy loads are considered in Chapter XIII. 

3. Repeated loads are forces that are applied a very large num¬ 
ber of times causing a stress (or stresses) in the material that is 
continually changing, usually through some definite range. 
For example, the loads applied to the connecting rod of an engine 
when the engine is running, the wheel loads on a railroad rail as a 
train passes over the rail, etc., are repeated loads. Repeated loads 
are discussed in Chapter XIV. 

Other Classifications of Loads. —Loads may be classified as dis¬ 
tributed loads and concentrated loads. A distributed load may be 
uniformly distributed or non-uniformly distributed. Thus, if 
sand be spread on a floor so that its depth is constant, the 
floor will be subjected to a uniformly distributed load, whereas, 
if the sand be distributed so that its depth is not constant the 
floor is said to carry a non-uniformly distributed load. A con¬ 
centrated load is one whose area of contact with the resisting body 
is negligible in comparison with the area of the resisting body. 

With reference to the manner in which the stresses in a body 
will vary from point to point and also to the general kind or type 
of stress and deformation that will be developed, the loads will, 
for convenience, be considered under three headings; namely, 
central loads, torsional loads, and bending loads. A body, how¬ 
ever, may be subjected simultaneously to loads of any two or to 
all three of these types. 

The stresses and strains caused by static central loads are dis¬ 
cussed in this chapter. The stresses and strains caused by static 
torsional loads are considered in Chapter IV, and those produced 
by static bending loads in Chapter V. 


STRESSES DUE TO CENTRAL LOADS 


5 


Static Central Loads 

3. Stresses Due to Central Loads.—A central load is a con¬ 
centrated load whose action line passes through the centroid of 
the area on which the stresses are to be considered, or a distrib¬ 
uted load whose resultant passes through the centroid of the area. 
The stress produced by a central load may be any one of the 
three types, tensile stress, compressive stress, or shearing stress, 
but the distinguishing feature of a central load as compared with 
torsional and bending loads is that the stress may be assumed to be 
uniformly distributed over the area, that is, the intensity of stress 
(unit-stress) may be assumed to be constant. If a central load 
acts normal to the area it is called an axial load and it may be a 



Tensile stres? Compressive stress Shearing' stress 

Fig. 1.—Tensile, compressive and shearing stresses. 


tensile or a compressive axial load according as it produces tensile 
or compressive stress; if the central load lies in the plane of the 
area it is called a shearing central load. 

Stresses due to central loads may be found as follows: Fig. 
1(a) represents a straight bar AB subjected to an axial tensile 
load P causing tensile stress on any cross-section of the bar; 
Fig. 1(c) represented a bar in compression under the action of an 
axial load P causing compressive stress in the bar; and Fig. 1(e) 
represents a body under the action of a central shearing load P 
causing shearing stress on the area A BCD. The problem is to 
find the relation between the load and the unit-stress developed 
for each of these three types of central loads. Let a section be 
passed through the body in each case and a free-body diagram for 
either part of the body be drawn, as shown in Figs. 1(6), 1(d), 
and 1(f). Since the load in each case passes through the centroid 



























6 


STRESS AND STRAIN 


of the area, the intensity of stress may be assumed to be constant; 
that is, the stress may be assumed to be uniformly 2 distributed 
over the area, and hence the total stress (called the resisting stress) 
is equal to the product of the area, a, and the unit-stress, s. There¬ 
fore, since each part of the body, in each case, is in equilibrium 
under the action of two collinear forces we may write 


Hence, 


load = resisting stress. 

P = a t St\ P = a c s c ; and P = a s s s , . . . . (1) 


in which the subscript denotes the kind of stress (tensile, com¬ 
pressive and shearing). Thus, in general, a 
central load, for any section of a body, is equal 
to the product of the area of that section and the 
unit-stress on the area. 

It should be noted that the load P of Fig. 
1(e) is not a central load unless the plate E is 
exceedingly thin, but in engineering computa¬ 
tions shearing loads may frequently be assumed, 
without serious error, to be central loads even 
though they do not comply strictly with the 
above definition of a central load. Further, the 
load with respect loads P of Fig. 2 are axial loads for section ran 
only 0116 section but not for any other section, and if the loads 
were applied at points Oi they would not be 
axial loads for any section of the body. 


Fig 



ILLUSTRATIVE PROBLEM 

Problem 1 . In determining the strength’ of concrete, a test cylinder 
8 in. in diameter and 16 in. high is loaded in a testing-machine as shown 
in Fig. 3(a). (a) If the maximum axial load Q that the concrete specimen can 

resist is 100,000 lb., determine the maximum unit-stress developed in the 
concrete. The dimensions shown in Fig. 3(6) and 3(c) were obtained from 
a testing-machine having a capacity of 100,000 lb.; when the concrete specimen 
is resisting its maximum load find (6) the maximum tensile unit-stress in 
each of the two screws of the testing-machine, (c) the shearing unit-stress in 

2 For a discussion of the influence of the relative dimensions of the body 
and of other conditions on the distribution of the stress over the cross- 
section, see Art. 136 to 139. 




STRESSES DUE TO CENTRAL LOADS 


7 


the threads of the screws, ( d ) the shearing unit-stress on the cylindrical 
area AB of the bronze bushing, and ( e ) the bearing unit-stress of the threads 
on the bushing. 

Solution. — (a) Since the load Q on the concrete specimen is an axial 
load the unit-stress on any section mn is uniformly distributed, and hence 
the total resisting stress is equal to a c s c (see Fig. Id). Hence, 

Q = u c s c , 

7T (8) 2 

100,000 = -y-.« c , 

s c = 1990 lb. per sq. in. 

Thus, the compressive stress in the concrete is 1990 lb. per sq. in. 



(6) A tensile load P of 50,000 lb. is resisted by each screw. Hence, 


P=a t s t , 

tt(2.35) 2 

50,000 = - 4 - s t , 

s t = 11,500 lb. per sq. in. 


The maximum tensile stress, then, that will be developed in the screws of 
this machine is 11,500 lb. per sq. in. 

(c) The area of the thread on which the shearing stress is developed as 
the thread resists being stripped from the screw is, 
















































































































8 


STRESS AND STRAIN 


Shearing area = a s = thickness of thread X circumference at root of 
thread X number of threads, n , in the depth, h, 

l h 

of the bushing in which the screw turns ( n — — f 
where p is the pitch) ' ^ 

1 6 

= -X7rX2.35X —= 22.15 sq. in. 

4 0.5 

Now since the load of 50,000 lb. on each screw is resisted by the shearing 
stress on this 22.15 sq. in. of shearing area, we have, 

P = a s s s , 

50,000 


22.15 


= 2260 lb. per sq. in. 


(d) The shearing area in the head of the bushing is 

a s = circumference X depth 
= 7 r 3.75 X2 = 23.55 sq. in. 

The shearing unit-stress on this area due to the 50,000-lb. load is 

P 50,000 ^ „ 

s 5 = - = 7^rr = 212 ° lb - P er S( h in * 
d§ 23 .55 

(e) The bearing area, a*>, of the threads on the bushing is 

ab = -^6 Xir 2.35 X12 = 16.6 sq. in. 

The bearing unit-stress, then, is 

P 50,000 

Sb = — =-= 3010 lb. per sq. in. 

db 16.6 

PROBLEMS 

2. If a specimen of wood is tested as shown in Fig. 4, and the maximum 
tensile unit-stress that the specimen can resist is 8000 lb. per sq. in., (a) 
what is the maximum axial load P that can be applied to the specimen? 
( b ) What is the shearing unit-stress in the heads of specimen when the 
load P is applied? 

Wall 



Fig. 4. —Tension and shear in wood specimen. 


Fig. 5.—Bearing pressure 
of washer on wall. 


3. A tie-rod 2 in. in diameter (Fig. 5) is used to help resist the lateral 
pressure against the walls of a bin. If the tensile unit-stress in the tie-rod 




























STRESSES DUE TO CENTRAL LOADS 


9 


is 15,000 lb. per sq. in., what diameter, d, should the washer have in order 
to keep the bearing unit-stress of the washer on the wall from exceeding 
200 lb. per sq. in.? Ans. 17.5 in. 


4. A timber frame shown in Fig. 6 carries a load, Q, of 12,000 lb. Find 
(a) the compressive unit-stress in the members A and B; ( b ) the shearing 
unit-stress in the timber C ; and (c) the compressive (or bearing) unit-stress 
of C on the blocks D, which are 4-in. cubes. 




Fig. 7.—Pin-connected wall bracket. 


5. The wall bracket shown in Fig. 7 carries a load, Q, of 10 tons. Find 
the tensile unit-stress in each of the two eye-bars, and the shearing unit- 
stress in the pins A and B if the diameter of each pin is 1 \ in. 

Ans. s* = 12,900 lb. per sq. in. s s = 7310 lb. per sq. in. 


6 . The bearing unit-stress for the collar bearing shown in Fig. 8 is 
90 lb. per sq. in., and the compressive unit-stress in the shaft is 6000 lb. 
per sq. in. If the diameter, di, of the shaft is 6 in., what is the load, P, on 
the shaft and the diameter, dz, of the collar? If the thickness, i, of the 
collar is 1 \ in., what is the shearing unit-stress on the area of contact between 
the collar and shaft? 


JLl 



ff 


Tv 

i 


Collar 



Fig. 8.—Collar 
bearing. 


Fig. 9. —Shear in key connect¬ 
ing pulley and shaft. 


7. A pulley (Fig. 9) transmits a turning moment, PXd, of 12,000 lb.-ft. 
to a 4-in. shaft, relative motion between the pulley and the shaft being 
prevented by a Hat key 1 in. wide, f in. deep, and 6 in. long. Compute 
the shearing unit-stress in the key. Ans. s s = 12,000 lb. per sq. in. 





































10 


STRESS AND STRAIN 


4. Strains Due to Central Loads .—Tensile Strain .—If an axial 

tensile load is applied to a straight bar of constant cross-section 
and of homogeneous material the bar is elongated or stretched 
and the strain per unit of length, that is, the unit-strain (unit 
elongation) e t is given by the expression 


in which e t is the total tensile deformation and l is the original 
length of the bar. If the cross-section of the bar is not constant 
or if the material is not homogeneous, all unit lengths of the bar 
will not elongate the same amount, and the above expression then 
represents only the average unit-strain; thus, since the unit-strain 
varies from section to section along the bar, the unit-strain at any 
section of the bar is the ratio of the elongation de t , of an elemental 
length dl, including the section, to the length dl. That is, the 
unit-strain at any section of the bar is 



(3) 


In order to determine e t from the above expression e t of course, 
must be expressed in terms of l (see the solution of Problem 12 
where this is done). 

Compressive Strain. —Similarly, if an axial compressive load is 
applied to a straight bar of constant cross-section and of uniform 
material (see Fig. lc) the bar is deformed (shortened) an amount 
e c and the unit-strain (unit-shortening) e c is 



(4) 


In loading a physical bar, it is practically impossible to obtain 
an axial compressive load; further, physical bodies are never 
homogeneous nor straight. Therefore, a compression member if 
relatively slender will bend when subjected to a load that is 
assumed to be axial. Bending action in compression members is 
discussed in Chapter XI. 

Shearing Strain .—Shearing strain (sometimes called detrusion) 
usually occurs in combination with tensile and compressive strain 
in connection with twisting and bending action, and it usually 
varies from point to point in the body. Although shearing strain 





STRAINS DUE TO CENTRAL LOADS 


11 


due to a central shearing load cannot be found experimentally as 
can tensile and compressive deformations, the meaning of shearing 
strain and the quantitative measure of shearing unit-strain 
will be discussed at this point. Thus, let Fig. 10(a) represent 
a bolt subjected to the shearing load P which tends to shear 
off the head of the bolt. The rectangle A BCD of Fig. 10(6) repre¬ 
sents a vertical section (enlarged) of that portion, A BCD, of the 
bolt that is subjected to shear. The head of the bolt exerts shear¬ 
ing stresses on the face BC and the lower part of the bolt exerts 
opposite shearing stresses on the face AD. Other forces would 
have to act on A BCD to hold it in equilibrium (that is P is not, 
strictly speaking, a central shearing load for section AD), but only 
the shearing forces will here be considered. The shearing forces 

shown in Fig. 10(6) cause the 
rectangle A BCD to assume the 
from of a rhombus AB\C\D. The 
part ABCD of the bolt sub¬ 
jected to shear may be considered 
to be made up of thin layers of 
materials each layer sliding a 
small amount relative to the layer 
beneath it. The total sliding, that 
is, the total shearing strain in 
the length A B or l is e s , and the shearing deformation per unit, 
length, that is, the shearing unit-strain e s , is 



/Ci 


Fig. 10.—Shearing strain. 


but 



tan 


where 0 is the change in the inclination of a line in the body that 
was originally perpendicular to the direction of the shearing 
deformation. Further, for a ,small angle, the tangent of the 
angle and the angle (expressed in radians) may be assumed to be 
equal without introducing serious errors. Therefore, 

The shearing unit-strain at any point in a body is measured 
by the change in inclination (expressed in radians) of two lines 
that pass through the point and that were originally at right angles 












12 


STRESS AND STRAIN 


Hence 

e s = y = tan 0=0.. (5) 

It will be shown later that a shearing stress (and also strain) on a 
given plane at a point in a body requires that there be a shearing 
stress (and strain) of equal intensity at the same point on a plane 
at right angles to the first plane, as is suggested by the distortion 
of the small cube within the body ABCD as shown in Fig. 10(6). 

It will be noted that a unit-strain (whether tensile, compressive, 
or shearing) is a length divided by a length and hence is an abstract 
number when each length is expressed in the same units; that is, 
its value is the same whether expressed as inches per inch or as 
feet per foot, etc. 

PROBLEMS 

8. A straight bar 6 ft. long and § in. in diameter is turned down to a 
diameter of \ in. for a distance of 2 ft. in its central portion. An axial 
load P causes a unit-deformation of 0.001 in the central 2 ft. and a total 
stretch of 0.04 in. in the whole bar. What is the unit-deformation of each 
of the end portions? 

9. From compressive tests of concrete cylinders it has been found that 
the concrete fails when the unit-deformation is about 0.0012. How much 
does a specimen 8 in. in diameter by 16 in. high shorten before failure occurs? 

5. Stress-strain Curve. Proportional Limit. Yield-Point. 
Ultimate Strength. Elastic Limit. Modulus of Elasticity.— 

Experiments have shown that for nearly all structural materials 
the unit-stress in a material is approximately proportional to the 
accompanying unit-strain of the material provided that the unit- 
stress does not exceed a certain value. For example, let a steel 
bar (Fig. 11) of length l and of constant cross-section, a, be sub¬ 
jected to an axial tensile load that gradually increases from zero 
value until the bar breaks. If P denotes any value of the load, 

P 

then the unit-stress, s t , corresponding to that load will be —. 

Further, let measurements of the stretch, e, of the bar be taken 
(by means of an extensometer) for various values of s t ; the unit- 

strain e, then, may be found from the expression € = y The rela¬ 
tion between the unit-stress, s t , and the unit-strain e, found experi¬ 
mentally as indicated above, is represented, within quite approx- 


STRESS-STRAIN CURVE 


13 


imate limits, by the stress-strain graph in Fig. 12(a), if the material 
is ductile such as low-carbon steel and other ductile metals, and 
by the stress-strain graph in Fig. 13 if the material is brittle, 3 such 
as high-carbon steel, etc. 


m 




Fig. 11.—Tensile 
test specimen. 



Proportional Limit .—As indicated in Figs. 12 and 13, it is 
found that for most structural materials, as the unit-stress is 
increased the unit-strain is increased in practically the same ratio; 
if the unit-stress is doubled the unit-strain is likewise doubled, 
etc.; that is, the stress-strain curve is a 
close approximation to a straight line 3 until 
the unit-stress reaches a value called the 
proportional limit (or limit of proportion¬ 
ality). This unit-stress is represented on 
the stress-strain curve by the ordinate to 
the point P-L in Figs. 12 and 13. There¬ 
fore, the proportional limit of a material is 
defined to be the maximum unit-stress that 
can be developed in the material without 
causing the unit-strain to increase at a 
greater rate than the unit-stress increases, 
as load is applied to the body. 

Yield Point .—As the load on the bar is increased further, 
causing a stress greater than the proportional limit, a unit-stress 



Fig. 13. — Stress-strain 
diagram for brittle 
material. 


3 The stress-strain curves for concrete and cast iron are curved practically 
all the way, but the first part of the curves, corresponding to relatively 
small stresses, may be assumed without serious error to be straight lines. 














14 


STRESS AND STRAIN 


is reached at which the material continues to deform without an 
increase in load, provided that the material is ductile. The unit- 
stress at which this action occurs is called the yield-point, and is 
represented on the stress-strain curve by the ordinate to the part 
CD of the graph in Fig. 12 (only ductile metals have yield-points). 
Thus the yield-point of a material is defined to be the unit-stress 
in the material at which the material deforms appreciably without 
an increase of load. 

The value of the unit-deformation for a bar when stressed to its 
proportional limit depends on the kind of material; the value for 
structural steel is about 0.0012, and for timber about 0.0020. 
Thus, if the bar in Fig. 11 were structural steel and its length, l, 
were 10 in., the total stretch e when the bar is stressed to its pro¬ 
portional limit would be e= el = 0.0012 X 10 = 0.012 in. Thus, 
in order to detect and measure the small strains that accompany 
stresses less than the proportional limit a strain-measuring appa¬ 
ratus (extensometer) is attached to the specimen. Further, the 
load is applied and measured by means of a testing machine 
similar to that shown in Fig. 3. The stretch that occurs while the 
bar is stressed at its yield-point, however, is relatively large, being, 
in the case of structural steel, as much as 0.025 in. per inch of 
length; a bar 10 in. long, then, would stretch one-fourth of an 
inch without an increase in load; this stretch is represented by the 
portion CD of the curve in Fig. 12. 

Ultimate Strength .—If the load on the bar (Fig. 11) increases 
still further, the unit-stress and unit-deformation increase as indi¬ 
cated by the portion of the curve DEF (Fig. 12a) if the material 
is ductile, until the maximum unit-stress is reached, which is repre¬ 
sented by the ordinate to the curve at F and is called the ultimate 
strength. The ultimate strength for a brittle material is repre¬ 
sented by the ordinate to F in Fig. 13; -a brittle material breaks 
when stressed to the ultimate strength whereas a ductile material 
continues to stretch. Hence, the ultimate strength of a material is 
defined to be the maximum unit-stress that can be developed in 
the material, as determined from the original cross-section of the 
bar or specimen; the cross-section of the bar decreases somewhat 
as the bar is stressed above the yield-point. 

After the ultimate strength of a ductile material is developed, 
the bar begins to “ neck down,” thereby rapidly reducing the area 
of cross-section at the neck-down section (Fig. 14), and the load 


STRESS-STRAIN CURVE 


15 


required to cause the bar to continue to stretch decreases, as indi¬ 
cated by the curve FG (Fig. 12a). The load on the bar at the 
instant of rupture is called the breaking load. The breaking load 
divided by the area of the neck-down section is the value of the 
unit-stress in the bar when rupture occurs and this value is con¬ 
siderably greater than the ultimate strength. The ultimate 
strength, however, is of more importance than any stress in the 
bar after necking down has started, since the bar is in the process 
of failing after necking down starts. 

Elastic Limit .—If the load on the bar is released after the 
bar has been stressed beyond the yield-point, the bar will not 


Neck-down 

Section 


Eig. 14. — Form of 
ruptured specimen 
of ductile steel. 

return to its original length, but will retain a part of its deforma¬ 
tion. The deformation per unit of length retained by the bar 
after the load (and stress) has been reduced to zero is called the 
'permanent set or merely set. For example, let the bar (Fig. 11) be 
stressed to the unit-stress, s, represented by the ordinate, ME, to 
the point E on the stress-strain curve in Fig. 15, the unit-strain 
corresponding to the unit-stress s being OM. If the load is grad¬ 
ually released the stress-strain curve will be represented by 
the line EN. That is, part of OM, represented by MN, is 
recovered but a part, represented by ON, is retained by the bar. 
ON therefore, represents the set corresponding to the unit-stress s. 

If however, the bar were subjected to a unit-stress equal to 


W 

A 



Fig. 15.—Stress-strain diagram showing perma¬ 
nent set. 

















16 


STRESS AND STRAIN 


(or less than) the unit-stress called the elastic limit, the bar will 
return to its original length when the load (and stress) is reduced 
to zero. Hence, the elastic limit of a material is defined to be the 
maximum unit-stress that can be developed in the material without 
causing a permanent set. 

In order to determine experimentally the value of the elastic 
limit for a material, a bar is subjected to a relatively small axial 
load and the load is released; if the extensometer shows that the 
bar has acquired no permanent deformation a larger load is 
applied and then released, etc. Thus, the stress in the bar is 
increased in small increments until a unit-stress is found which, on 
release of the stress, leaves a very small set in the bar, the minimum 
unit-stress at which set first occurs being the elastic limit. The 
results of tests show that for most structural metals the elastic 
limit of the metal has approximately the same numerical value as 
has the proportional limit of the material, and in technical liter¬ 
ature, the proportional limit frequently is called (though incor¬ 
rectly) the elastic limit. Further, the yield-point is sometimes 
called the commercial elastic limit. 

Modulus of Elasticity. —If the bar discussed in the preceding 
article were subjected to a compressive stress, or to a shearing 
stress instead of a tensile stress, the stress-strain curve would be of 
the same form as that shown in Figs. 12 and 13. Hence, it follows 
that when a material is stressed in one direction 07ily, the unit- 
stress at any point in a material is proportional to the unit-strain 
at that point, provided that the unit* stress does not exceed the 
proportional limit of the material. That is, for stresses within the 
proportional limit, all elastic material behaves according to the 
same law; namely, that the ratio of the unit-stress to the unit- 
deformation is a constant, or expressed mathematically, 

s 

- = a constant,.(6) 


regardless of the kind of stress developed (whether tensile, com- 
presive or shearing) and regardless of the way the stress and defor¬ 
mation are produced (whether by axial, bending or torsional loads, 
etc.). This is known as Hooke’s law. 4 The numerical value of 

4 As pointed out earlier in this article, this law should be regarded as 
approximate, the error involved being negligible for most computations 
involving static loads and ductile materials but not necessarily negligible 
when repeated loads are involved, as will be discussed in Chapter XIV. 



MODULUS OF ELASTICITY 


17 


this constant ratio, however, is in general different for any one 
material when subjected to the different types of stress, and is also 
different for the different materials subjected to the same types of 
stress. 

The numerical value of the constant ratio of the unit-stress in 
a material to the accompanying unit-strain within the propor¬ 
tional limit is called the modulus of elasticity of the material. The 
symbol E will be used to denote the modulus of elasticity, and sub¬ 
scripts t, c and s will be used to denote tensile, compressive and 
shearing, respectively. Thus, we may write 

E,=~, E c =~ and E s =- .(7) 

€ c e s 


and, when the stress and deformation are caused by central loads, the 
above expressions may be written as follows: 


F - a - pi \ F - Pl 

JtLt -—> - 

a i CL6t ae c 

i 


, p a P 
and E s = — = —- 
c s dcf) 

i 


. ( 8 ) 


The value of E s is not found experimentally, however, from the 
above expression since the shearing deformation <f> (see Art. 4) 
caused by a central shearing load is very difficult if not impossible 
to measure. The value of E s , however, can easily be found when 
the shearing unit-stress and shearing unit-strain are produced by 
torsional loads as is discussed in Chapter IV. 

It should be noted that the modulus of elasticity is expressed 
in the same units as is unit-stress since e is a ratio of length to 
length and is therefore merely a number. 

As stated above, the values of the moduli of elasticity for 
any one material are, in general, not the same. For steel the ten¬ 
sile and compressive moduli of elasticity, E t and E c , are approx¬ 
imately 30,000,000 lb. per sq. in., but the shearing modulus, E s , is 
approximately 12,000,000 lb. per sq. in. Similarly, any one of 
the moduli has different values for different materials. Thus, 
average values of the tensile moduli for steel, cast iron and timber 
(pine) are respectively 30,000,000, 15,000,000 and 1,500,000 lb. per 
sq. in. 


18 


STRESS AND STRAIN 


It follows from the definition that the value of the modulus of 
elasticity of a material is represented by the slope of that portion 
of the stress-strain curve below the proportional limit; this portion 
of the curve is usually drawn to a large scale as in Fig. 126 so that 
the value of the slope can be obtained with reasonable accuracy 
(see also Art. 144). Now this slope represents the rate at which 
unit-stress increases with unit-strain and hence the modulus of 
elasticity of a material is a measure of the stiffness of the material. 
That is, if one material has a modulus of elasticity twice as great 
as another material, the resisting stress in the one material, for a 
given strain, is twice as great as that in the other, and hence the 
one material is twice as stiff as the other. 

6. Properties of Structural Materials. —The proportional limit, 
elastic limit, yield-point, ultimate strength, modulus of elasticity, 
etc., of a material are usually called physical (or mechanical) 
properties of the material, the values of which are found from 
experimental results, and the numerical values of the properties 
are frequently called physical constants. Much of the material 
used in structures and machines is bought according to specifica¬ 
tions which give the values of the physical properties that the 
material shall have for various uses. A number of technical 
societies and engineering companies have published specifications 
of structural materials; perhaps the most complete specifications 
are those of the American Society for Testing Materials called 
the “ A.S.T.M. Standards.” 

Average values of various properties for a few of the more 
common engineering materials are given in Table 1. Attention is 
called to the fact that the properties of a commercial material 
necessarily vary somewhat due to the uncontrollable factors in the 
method of manufacture, treatment, etc., and that the specifica¬ 
tions usually make reasonable allowance for this variation by 
specifying a range of values or a minimum value for the properties 
which a material must have. For example, the ultimate strength 
of “Structural Steel for Building” as given in the “A.S.T.M. 
Standards ” is 55,000 to 65,000 lb. per sq. in. and the yield-point 
must be not less than one-half of the ultimate strength. 

The average values in Table 1 are presented with the view 
of helping the student to develop judgment in the use of engi¬ 
neering materials; the values in the table should be used when 
needed in the solution of subsequent problems. A detailed dis- 


TABLE 1 

Average Values of Strength, Stiffness, and Ductility of Various Structural Materials 


PROPERTIES OF STRUCTURAL MATERIALS 


19 


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r)<*0<N 




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«-c 


(6) Not well defined. 

(c) Somewhat greater than the tensile strength since the material fails in tension when tested in shear. 



































20 


STRESS AND STRAIN 


cussion of the significance of the properties of structural mate¬ 
rials is given in Chapter XV, and tables giving values of the prop¬ 
erties of most of the structural materials are given in Chapter XVI. 

In commercial testing of ductile material, such as structural 
steel, the four quantities usually found are the yield-point, the ulti¬ 
mate strength, the elongation in per cent and the reduction of area 
in per cent; the proportional limit, elastic limit and modulus of 
elasticity are seldom found in commercial tests. 

The percentage of elongation is found by dividing the increase 
in length of a bar after rupture has occurred by the original length 
and multiplying by 100; and the percentage of reduction of area 
is found by dividing the difference between the areas of the rup¬ 
tured and original sections by the area of the original section and 
multiplying by 100. The percentage of elongation and of reduc¬ 
tion of area are measures of the ductility of a material (see Art 
141 for further discussion). 

The following facts should be noted from a study of Table 1: 

(а) The shearing proportional limit for steel is about six-tenths 
(0.6) of the tensile proportional limit. 

(б) The ultimate strengths of the more ductile steels are about 
twice as large as the yield-points. 

(c) The tensile moduli of elasticity of all grades of steel are 
equal, the value of the modulus being 30,000,000 lb. per sq. in.; 
thus the stiffness of steel is constant, whereas the strength of steel 
varies greatly with the composition and treatment. 

(d) The shearing moduli of elasticity of all grades of steel are 
equal, the value of the modulus being two-fifths of the tensile 
modulus; E s = % E t = % 30,000,000= 12,000,000 lb. per sq. in. 

(e) The maximum useable compressive strength of ductile 
material is the yield-point of the material. 

(/) The compressive strength of brittle material (cast iron, 
concrete, stone, etc.) is greater than the tensile strength. 

(i g ) The strength of timber varies greatly with the direction of 
the grain. 

ILLUSTRATIVE PROBLEMS 

Problem 10. A short concrete compression member is reinforced with 
12 rods of steel % in. in diameter arranged in a circle with a 5-in. radius as 
shown in Fig. 16. A uniform pressure or load is applied to each end surface 
of the member. The concrete outside of the reinforcing rod is used for pro¬ 
tecting the rods in case of fire and is not assumed to resist any of the load. 


PROPERTIES OF STRUCTURAL MATERIALS 


21 


If the load causes a compressive unit-stress of 600 lb. per sq. in. in the 
concrete, what is the unit-stress in the steel rods, and what is the total load 
carried by the member. The compressive moduli 
of elasticity of the concrete and steel are Ei = 2,000,- 
000 lb. per sq. in., and ^2 = 30,000,000 lb. per sq. 
in., respectively. 

Solution .—Let the subscript 1 refer to concrete 
and the subscript 2 refer to the steel. Thus, 


and 


Ei=— or Si=Eiei . (9) 

ei 


E -2 = — or s 2 =E 2 e 2 . . (10) 

€2 



But since the concrete and the steel shorten Fig. 16. —Concrete com- 

the same amount, ei = e*. Hence, dividing (9) by pression member rein- 
, ” forced with steel rods. 

(10), we have 

$ 2 E« 

- = — = 15, 

si Ei 


or 


s 2 = 600X15 = 9000 lb. per sq. in. 


Therefore, the compressive stress in the steel is 9000 lb. per sq. in. Thus, 
the stress in the steel increases with deformation 15 times as fast as does 
the stress in the concrete; that is, steel is much stiffer than is concrete. 

The total load P, then, is 


7r(i) 2 tt(10) 2 

P = 9000 X12 X —^—|-600X“—— 




/, 



= 68,300 lb. 

Problem 11.—A bar is l in. long, has a constant cross- 
sectional area of a sq. in., and weighs w lb. per ft. of length 
per sq. in. of cross-section. Find the total elongation of 
the bar when it is suspended from one end and is sub¬ 
jected to no downward load except its own weight. If the 
bar is made of steel and is 400 ft. long, calculate the total 
stretch of the bar. (A steel bar having a cross-sectional 
area of 1 sq. in. weighs 3.4 lb. per ft. of length.) 

Solution .—The unit-stress, s*, on a section at any dis¬ 
tance y in. (Fig. 17) from the lower end expressed in lb. per 
sq. in. is 


Fig. 17.—Bar 

stretched by g _ awy _wy 
its own weight. 1 12 a 12 


and the unit-deformation at this section is 























22 


STRESS AND STRAIN 


The elongation of the short length, dy in., of the bar along which the unit- 
elongation may be assumed to be constant is 


de t =e t dy, 


and the total elongation (in inches) in the length of l in. is 



3.4X(400X12) 2 

12X30,000,000X2 


= 0.107 in. = elongation of steel bar 400 ft. long. 


PROBLEMS 

12 . It is specified that a steel rod 40 in. long is to be subjected to a 
unit-stress not greater than 10,000 lb. per sq. in. and to be elongated not 
more than 0.01 in. when resisting a tensile axial load of 20,000 lb. Deter¬ 
mine the cross-sectional area required to satisfy each of the specifications 
and state which requirement governs the design. 

13. Load in Reading of Exten- 


Pounds 

someter in Inches 


500 

0.0 


1500 

0.0005 


500 

0.0 

In a tension test of a steel bar 

3000 

0.0010 

0.499 in. in diameter the elongation 

500 

0.0 

was measured over a gage length of 

4490 

0.0020 

2 in. Successive readings of the load 

500 

0.0001 

and of the extensometer were as given 

5980 

0.0025 

herewith. Determine the elastic limit, 

500 

0.0002 

the proportional limit and the yield- 

7510 

0.0030 

point of the material. 

500 

0.0005 


8630 

0.0035 


500 

0.0010 


9500 

0.0010 


500 



9600 

0.0130 



14. Two blocks, each 4 in. by 8 in. by 40 in., are bolted together to form 
a compression member as shown in Fig. 18. A pressure is applied to the 
top surface of the member causing both blocks to shorten the same amount. 




WORKING STRESS 


23 


If one of the blocks is made of gray cast iron and the other of oak, find the 
total load P when the unit-stress in the cast iron is 25,000 lb. per sq. in. 


Cover plate 






Oak 

Cast 

Iron 



<—4—> 

<—4^-> 


i 




Fig. 18. —Two-material 
compression block. 



15. Two wires, one of steel for which E t is found to be 14,500 tons per 

sq. in., and one of copper for which E t is found to be 7500 tons per sq. in., 
have the same length and carry equal axial loads. The copper wire has a 
diameter of 0.03 in. If each wire elongates the same amount, what is the 
diameter of the steel wire? Ans. d = 0.0155 in. 

16. Will a permanent set be caused in a wrought-iron bar f in. in diameter 
and 3 ft. long when subjected to an axial tensile load of 24,000 lb.? 

17. A steel piano wire with a constant cross-section of 0.00038 sq. in. 
has an elastic limit of 100,000 lb. per sq. in. If the wire is used to let down 
a 30-lb. body from the top of a building 500 ft. high, what must be the 
original length of the wire if the body just touches the ground when sus¬ 
pended by the wire? (Steel weighs 0.28 lb. per cu. in.) 

18. The boiler brace shown in Fig. 19 resists the pressure on an area 
of 80 sq. in. of the boiler-head, (a) If the steam pressure in the boiler is 
120 lb. per sq. in., what is the tensile unit-stress in the brace? (6) If the 
brace is made of steel, how much does it elongate; assume that the rod is 
of constant diameter from pin to pin. (c) What is the shearing unit-stress 
in the three rivets at A if the diameter of each rivet is f in.? 

Ans. (a) 16,900 lb. per sq. in., (5) e = 0.0182 in., (c) s s = 6780 lb. per sq. in. 

19. How long must be the bar described in Problem 12 in order to cause 
a stress in the bar equal to the proportional limit of the material? The bar 
is made of structural steel. 


7. Working Stress. —A working stress or an allowable stress for a 
material is the maximum unit-stress that is considered to be safe 
for the material when the material is resisting the loads that are 
assumed to be applied to it in service. The values of allowable 



























24 


STRESS AND STRAIN 


stresses that are commonly used for various materials when resist¬ 
ing various types of loading have been determined largely from the 
results of tests of the materials and from the accumulated experi¬ 
ence obtained in the construction and use of structures and 
machines under service conditions; Table 2 gives commonly used 
values of working stresses for several materials when the material 
is subjected to static loads (see Table X of Chapter XVI for a 
more extensive table of values). 

The allowable working stress in a member is always consider¬ 
ably less than the ultimate strength of the material of the member. 
For brittle materials (materials that do not have yield-points), 
such as concrete, cast iron, high carbon steel, etc., the working 
stress is frequently taken as a certain proportion 5 of the ultimate 
strength of the material. For ductile materials, such as structural 
steel, wrought iron, etc., the working stress may be taken to be 
either a certain proportion 5 of the ultimate strength or a certain 
(but different) proportion of the yield-point. In general, the 
working stress for a ductile material must be less than the yield- 
point (and usually considerably less) since most structures would 
not fulfill their function if the members of the structure were 
stressed beyond their yield-points and thus became permanently 
deformed, although they might be safe against rupture or col¬ 
lapse. (For a further discussion of this point see Art. 138.) 

Need of a Margin of Strength .—The need for selecting working 
stresses considerably less than the ultimate strength (or less than 
the yield-point) arises from (1) the uncertainties as to the prop¬ 
erties of the materials used, (2) the uncertainties as to the loads to 
be resisted by the structure or machine as a whole and also by the 
various members of the structure, and (3) the uncertainties in the 
methods of calculating the stresses in the members. The more 
these uncertainties are reduced the higher the working stresses 
may be. And, if the working stresses are increased, the amount 

5 The term factor of safety of a member has been widely used to denote 
the ratio of the ultimate strength of the material of the member to the 
working stress used in designing the member. The use of this term and 
the idea it conveys, however, is now generally conceded to be misleading 
and hence undesirable. For example, if a factor of safety of 4 is used with 
structural steel in selecting a working stress, a structure designed on the 
basis of this working stress would not resist loads four times as great as 
the loads assumed to act on the structure and to produce the working stresses. 
For further discussion, see Arts. 138 and 139. 


25 


WORKING STRESS 


of materials purchased (and hence the cost of construction) will 
decrease. 

In building constructions, working stresses have become, to a 
large degree, standardized. Thus, the building laws of most large 


TABLE 2 

Values of Working Stresses for Use with Static Loads 
(See also Table X of Chapter XVI) 

Values of stresses are expressed in lb. per sq. in. 


Material 

Type of Stress 

Proportion of 

Direct 

Tension 

Direct 

Compression 

Direct 

Shear 

Ultimate 

Strength 

Structural steel. 

16,000- 

18,000 

14,GOO- 

16,000 

8 ,000- 

10,000 

( 0.25 for tension * 

10.20 for shear 

Wrought iron. 

15,000 

12,000 

8000 

/ 0.25 for tension * 

10.20 for shear 


Gray cast iron. 

3,000 

15,000 


( 0.15 foi tension 

10.20 for compression 

Timber (yellow pine). 


1000 with the 
grain 

250 across the 
grain 


0.15 with the grain 

Portland cement con¬ 
crete; 1:2:4 mix. 


450 

.... 

0.22 


* For ductile material, such as structural steel, w ought iron, etc., the working stress is 
usually thought of as a proportion of the yield-point since for such material the yield- 
point, rather than the ultimate strength, is the maximum useable strength of the material 
(see Art. 138); if this were done, the values would be 0.5 for tension and 0.4 for shear, since 
the yield-points of structural steel and wrought iron are approximately one-half of the 
ultimate strengths. 


cities specify the maximum allowable working stresses to be used 
for the common structural materials (see Table X of Chapter XVI). 
For example, a working stress of 16,000 lb. per sq. in. 6 is specified 

6 When structural steel is purchased according to standard specifications, 
and subjected to rigid inspection, etc., working stresses higher than 
16,000 lb. per sq. in. (18,000 to 20,000 lb. per sq. in.) are sometimes allowed 
n tension and flexural members; for steel compressive members, however, 
the working stress is usually specified to be less than 16,000 lb. per sq. in. 
(usually from 12,000 to 14,000 lb. per sq. in.; see Art. 139). 



















26 


STRESS AND STRAIN 


by most building codes for tension members and for beams, etc., 
when made of rolled structural steel. In machine construction 
standardization of working stresses is more difficult than in 
structural work since the variety of materials used is greater and 
the conditions of service are more uncertain. However, a manu¬ 
facturing firm that builds machines of certain types is able to 
standardize the working stresses for the design and construction 
of those machines. 

The decrease in cost of construction accompanying the use of 
higher working stresses acts as a stimulus to reduce the uncer¬ 
tainties referred to above. Thus, the methods of manufacture 
and treatment of materials are being refined; tests and inspection 
of materials are becoming more rigid and more generally used; 
knowledge of the properties of materials and of the methods of 
calculating stresses in the members is becoming more extended 
and more generally applied, and experience and tests are gradu¬ 
ally yielding more definite information as to the loads that various 
types of structures and machines are subjected to in service. 
The standardization of working stresses by law and by practice, of 
course, helps greatly to safeguard the public, who use engineering 
structures such as buildings, bridges, locomotives, elevators, etc., 
against unscrupulous or incompetent engineering design. 

Factors Affecting Values of Working Stresses .—From Table 2 
it will be noted that the working stresses for cast iron and timber 
are relatively small in relation to the ultimate strength of the 
material, the reason being that these materials are less uniform in 
structure than steel. For example, cast iron may contain blow 
holes, and initial stresses due to uneven cooling; timber may con¬ 
tain knots, pitch pockets, cracks, etc., and its strength is influ¬ 
enced markedly by its moisture content. Further, for a ductile 
material such as structural steel working stresses are relatively 
larger in relation to the ultimate strength than for a brittle mate¬ 
rial such as hard steel, cast iron, etc. The reason for this is that a 
ductile material deforms or yields if subjected to unexpected over¬ 
loads or if high localized stresses occur at some portion of a member 
(and such stresses always do occur); the yielding distributes some 
of the excess stress to the surrounding materials and hence tends 
to prevent the member from breaking. (See Arts. 138 and 143 for 
further discussion.) 

The working stress for any one material is, in general, smaller 


WORKING STRESS 


27 


when the material resists impact and repeated loads than when it 
resists static loads; the reasons are that an impact load is in gen¬ 
eral less definitely known and it causes larger stresses and strains 
than does the same load when applied gradually, and localized 
stresses and non-homogeneity of material have a much greater 
influence on the strength of the member when the loads are 
repeated than when applied but once as a static load (see Art. 119 
of Chapter XIII and Art. 131 of Chapter XIV for further discus¬ 
sion). 


PROBLEMS 

20. When a force, P, of 196 lb. is applied to the bell-crank shown in 
Fig. 20, the bearing pressure normal to the rubbing surfaces of the friction 
clutch is 15 lb. per sq. in. If the coefficient of friction for the two surfaces 
is 1, what is the maximum shearing unit-stress that can be developed in the 
rectangular key if its dimensions are 1 in. wide, f in. deep and 4 in. long? 
What is the ratio of this stress to the shearing yield-point of the material if 
the key is made of structural steel? (Tensile yield-point = \ tensile ultimate 
strength; shearing yield-point = 0.6 tensile yield-point.) 



nected by pin. 


21. In the joint shown in Fig. 21, di = 2\ in., d 2 = ll in., t=2 in., 6 = 1 in., 
and c = 5 in. All of the material is structural steel. If the load P is 50,000 lb. 
and if the working stress is specified to be not greater than 0.3 of the yield- 
point, does the joint satisfy the specification? (For values of yield-points, 
see statement in Prob. 20.) 
















































28 


STRESS AND STRAIN 


22. A hollow gray cast-iron compression member supports loads as shown 
in Fig. 22. The member is prevented from bending by lateral supports 
(not shown). If the outside diameter is 20 in., what should be the inside 
diameter if the working stress is one-eighth (§) of the ultimate strength? 


80,000 lb. 



Fig. 22. —Hollow com¬ 
pression member. 



23. The capacity of the hook and yoke shown in Fig. 23 is 10 tons. If 

the material has a yield-point of 40,000 lb. per sq. in. in tension and six- 
tenths as large in shear, what are the ratios of the tensile stress at section A, 
and of the shearing stress in the pin to the corresponding yield-points when 
the hook is subjected to its maximum load? Ans. 0.24; 0.34. 

24. Figure 24 represents a lower panel point of a pin-connected Pratt 
truss. The post and the three eye-bars are made of structural steel having 
an ultimate strength of 65,000 lb. per sq. in. and a yield-point equal to 
one-half the ultimate strength. Find the area of each of the eye-bars using 
a working stress equal to 0.4 of the yield-point. 

Ans. up = 2.18 sq. in., gq = 3.08 sq. in. 


25. Figure 25 represents two bars of diameter d\ joined by means of a 
cotter pin, the axes of the bars being in the same straight line. The upper 
bar is enlarged at its lower end to form a hollow cylindrical socket which 
fits over the enlarged upper end of the lower bar. The two bars and the 
































WORKING STRESS 


29 


cotter pin are made of structural steel. From the following data find the 
ratios of the tensile stress in the two bars, of the tensile stress in the socket, 
of the shearing stress in cotter pin, and of the bearing stress on the cotter pin 




Fig. 24.—Stresses in pin-connected truss. Fig. 25.—Socket joint 

and cotter pin. 

to the corresponding ultimate strengths of the material. The ultimate 
bearing stress may be assumed to be 80,000 lb. per sq. in. 

P = 60,000 lb.; di=2 in.; d 2 = 2\ in., d 3 = in., d 4 = in., t = 1 in., 
and h = 3 in. 












































CHAPTER II 


THIN-WALLED CYLINDERS AND SPHERES. RIVETED 

JOINTS 

8. Stresses in Thin-walled Cylinders and Spheres. —A thin- 

walled cylinder or sphere is one in which the thickness of the wall 
or shell is small in comparison with the diameter of the vessel. 
When this condition is satisfied the stress in the shell due to an 
internal fluid pressure may be considered to be uniformly dis¬ 
tributed on the cross-sectional area of the shell without intro¬ 
ducing serious errors in the calculation of the stress. Boilers, 
tanks, water and steam pipes usually may be treated as thin- 
walled cylinders, and hoops and tires that are shrunk on wheels 
may also usually be so treated. The cylinders of large guns and 
of some pipe, such as the pipes to hydraulic forging presses, etc., 
have thick walls and the stress in the walls cannot be assumed to be 
uniformly distributed without introducing a large error in the cal¬ 
culated unit-stress. 

The problem here considered is to determine the relation 
between the internal pressure in a closed thin-walled cylinder, 
the diameter of the cylinder, the thickness of the shell, and the 
intensity of stress in the shell (a) on a longitudinal section and (6) 
on a transverse section. 

Stress on a Longitudinal Section. —Fig. 26(a) represents a por¬ 
tion of a thin-walled cylinder that is subjected to an internal 
fluid pressure of intensity R, the length of the portion being l. 
Let the diameter of the cylinder be denoted by D, the thickness of 
the shell by t, and the intensity of the tensile stress in the shell 
by s t . 

The pressure of the fluid on the internal surface of the cylinder 
at any point is normal to the surface at that point, as indicated in 
Fig. 26(a), and these internal pressures tend to rupture the cylinder 
on a longitudinal section (at AB and EF, Fig. 26a), the resultant 

30 


STRESSES IN THIN-WALLED CYLINDERS AND SPHERES 31 


pressure (or load) on one-half of the shell (Fig. 26 b) being resisted, 
and held in equilibrium, by the total stresses P, P exerted by the 
other half of the shell at the areas AB and EF. Hence from 
equilibrium we have, 

Resultant horizontal pressure or load = total resisting stress. 

An expression for the resultant horizontal pressure on the half¬ 
shell may be found as follows: The resultant horizontal pressure 
on the semi-cylindrical area, a, of the half-shell (Fig. 266) is the 
sum of the horizontal components of the pressures on the ele- 



Fig. 26 .— Stress on longitudinal section of thin-walled cylinder. 


mentary areas; the pressure on an elementary area da is Rda and its 
horizontal component is Rda cos 0, and hence the resultant horizon¬ 


tal pressure is equal toj ^Rda cos 0,which may be written, rJ< 


da cos 0 


since the pressure R is the same at all points in the semi- 
cylindrical area, a. But da cos 0 is the area formed by projecting 

the area da on a vertical plane and hence Jda cos 0 is the area 

formed by projecting the semi-cylindrical area on a vertical plane, 
and is, therefore, equal to Dl. Thus, the resultant horizontal 
pressure is Rdl. 

Further, since the shell is thin, the total resisting stress, 2 P, 
may be assumed to be distributed uniformly over each of the two 























32 


THIN-WALLED CYLINDERS 


areas, and hence 2P = 2a t s t = 2ltst. Therefore, the above equation 
becomes 


Whence, 


RDl = 2lts t . 


RD 
S,= ~2t- 


( 11 ) 


Stress in Transverse Section .—The total pressure of the 
fluid against the end of the cylinder must be resisted by 



Fig. 27.—Stress on traverse section of 
thin-walled cylinder. 


the total stress on a trans¬ 
verse section of the cylinder 
as indicated in Fig. 27. Now 
the total pressure against the 

end of the cylinder is R 

and the total resisting stress 
is 7 rDt-s t . Hence from the 
condition of equilibrium we 
have, 


whence, 


R^=*Dt 

4 






RD 
4 1 ‘ 


( 12 ) 


In (11) and (12) R and s t must be expressed in the same units 
(usually pounds per square inch), and 
D and t must be expressed in the same 
units (usually inches). 

A comparison of equations (11) and 
(12) shows that the intensity of stress 
on a longitudinal section of a thin-walled- 
cylinder due to an internal fluid pres¬ 
sure is twice as great as that on a 
transverse section of the same cylinder. 

This fact explains why the riveted joint 
connecting the plates of a boiler along 
a longitudinal seam requires more rivets than that along a trans¬ 
verse seam. 

Equation (12) may be used also to find the unit-stress in a thin- 
walled sphere due to internal fluid pressure as will be evident 



Fig. 28.—Stress in thin- 
walled sphere. 


















STRESSES IN THIN-WALLED CYLINDERS AND SPHERES 33 


when the method used above in determining the stress in a thin- 
walled cylinder is applied to a thin-walled sphere as represented 
in Fig. 28. 


PROBLEMS 

26 . A boiler is subjected to an internal pressure of 80 lb. per sq. in. and 
is 48 in. in diameter, (a) What is the total stress acting on (or transmitted 
across) each of the two longitudinal cross-sectional areas of the plate or 
shell in a length of 3 in. along the boiler? (b) What is the unit-stress 
developed in the shell if the thickness of the shell is \ in.? 

27 . A standpipe 6 ft. in diameter is 60 ft. high. When it is full of water, 

what is the circumferential unit-stress in the plate at the bottom of the stand¬ 
pipe if the thickness of the plate is f in.? Hint: Since water weighs 62.5 lb. 
per cu. ft., the pressure (in all directions) at the bottom of the standpipe is 
62.5X60 lb. per sq. ft. Ans. s = 2500 lb. per sq. in. 

28 . The pressure in the cylinder of a steam-engine (Fig. 29) is 120 lb. 
per sq. in. and the internal diameter, D, of the cylinder is 14 in. How 
many f-in.bolts are required for strength 
if the tensile unit-stress is not to exceed 
8000 lb. per sq. in.? What should be the 
thickness, t , of the walls of the cast-iron 
cylinder to satisfy the requirement for 
strength if the allowable tensile stress is 
8000 lb. per sq. in.? Note. —The allow¬ 
able stresses are taken low due to the 
fact that the load is applied with more 
or less impact. Further, the requirement 
for strength in many problems is not the 
governing requirement. In this problem, 
for example, the bolts should be large 
enough to prevent a workman, with or¬ 
dinary tools, from twisting the heads off. Further, the requirement for tight¬ 
ness of the joint may determine the number of bolts. Similarly, the thick¬ 
ness of the wall may be influenced by the considerations of heat loss or of 
ease and reliability in casting, etc. 

29 . Water pipes are frequently made of cast iron. According to specifica¬ 

tions a pipe 18 in. in diameter must have a wall thickness of 0.87 in. and 
must resist an internal pressure of 300 lb. per sq. in. What circumferential 
unit-stress is developed in the pipe? Ans. s = 3100 lb. per sq. in. 

30 . The steel tire for a locomotive driving-wheel has an internal diameter 
TsV o ‘d l ess than that of the wheel on which the tire is to be shrunk, where 
d is the diameter of the wheel. The value of d is 60 in. and the value of t, 
the^thickness of the tire, is f in. If it is assumed that after the tire is shrunk 



Fig 29. —Steam cylinder and 
piston. 
















34 


THIN-WALLED CYLINDERS 


on the wheel the diameter of the wheel is not changed by the pressure of 
the tire, find (1) the elongation of the tire, (2) the tensile unit-stress (hoop- 
tension) in the tire, and (3) the intensity of the pressure of the tire on the 
wheel. Am; (1) e = 0.126in., (2) s = 20,000 lb. per sq. in., 

(3) R = 500 lb. per sq. in. 

Riveted Joints 

9. Introduction.—Riveted joints are important structural 
elements in buildings, bridges, cranes, etc., and also in pressure 
vessels such as boilers, tanks, water pipes, etc. 

The load causing the stresses in the various parts of a riveted 
joint (shearing stress in the rivets, bearing stress on the rivets, 
tensile stress in the plate, see Fig. 32) is usually assumed to act 
through the centroid of the area on which the stresses occur, and 
hence the stress on the resisting area is assumed to be uniformly 
distributed. This assumption, however, must be regarded as a 
rough approximation only, since, as discussed in Art. 13, the load 
on a riveted joint is seldom distributed evenly to the rivets; how¬ 
ever the design of the simpler types of joints are usually based on 
this assumption. The stresses in riveted joints produced by eccen¬ 
tric loads are discussed in Chapter X, but eccentric loads, when¬ 
ever possible, should be avoided. 

According to the above assumption the stresses in a riveted 
joint may be found from the equation developed in Chapter I for 
a central load; namely, P = as, in which P is the central load trans- 
mited from one to the other of two plates that are connected by 
the rivets, and s is any one of the various unit-stresses (shearing, 
bearing, tension, etc.) that may be the cause of the failure of the 
joint (as discussed in Art. 11) and a is the area on which the stress 
is distributed. 

10. Types of Riveted Joints. Definitions.—It is convenient 
to divide riveted joints into two general groups; (1) structural 
joints used in connecting members in bridges, buildings, cranes, 
and other structures, and (2) boiler, tank, or pipe joints used in 
connecting plates in various types of pressure vessels. In struc¬ 
tural joints, strength is the main requirement whereas in boiler 
and pipe joints tightness in addition to strength must be con¬ 
sidered. 

Two types of joints are widely used for both the groups men¬ 
tioned above; namely, lap joints and butt joints (see Fig. 30), 


TYPES OF RIVETED JOINTS—DEFINITIONS 


35 


and both lap and butt joints may be single riveted, double riveted, 
triple riveted, etc.,-according as one, two, three, etc., rows of 
rivets pierce each of the two plates that are connected. Further, 
a butt joint may have two cover plates (or straps) or only one 
cover plate. Again, both lap and butt joints may have the rivets 
arranged in the form of chain riveting (Fig. 306 and 30c) or in the 
form of staggered riveting (Fig. 30d and 30e). 

Definitions. —The terms defined below will be used frequently 
in the subsequent articles. 

The 'pitch of a row of rivets is the distance between the centers 



( f : 

-c!> 


N* 




(a) Single riveted lap joint (6) Double riveted lap joint (c) Single riveted butt joint 



(c?) Double riveted butt joint 



Fig. 30.—Types of riveted joint. 


of any two adjacent rivets in the row. The pitch is not necessarily 
the same for all rows. Thus, in Fig. 30c the pitch for the inner 
rows is denoted by p and for the other row by p\ in which pi 
equals 2 p. 

The transverse pitch is the distance between the center lines of 
two rows of rivets, denoted by p t in Fig. 30(d), and the diagonal 
pitch in staggered riveting is the distance from the center of a 
rivet in one row to that of the nearest rivet in the next row, 
denoted by p d in Fig. 30(d). 

The length of a repeating group of rivets is used in connection 
with boiler and pipe joints to denote the shortest distance along the 

























































































36 


THIN-WALLED CYLINDERS 


joint that includes a characteristic group of rivets which recurs 
along the length of the joint. The length of the repeating group 
is usually equal to the pitch, the maximum pitch being used when 
the pitches of all rows are not the same. In computations it is 
convenient to use the force transmitted through the repeating 
group of rivets rather than that transmitted through the entire 
length of the joint. 

The margin of a plate in a riveted joint is that part of the plate 
between the edge of the plate and the center line of the nearest 
row of rivets. 

A gusset plate or splice plate is a plate used in structural joints 
to form part of the joint in connecting two or more members of a 
structure (see Fig. 336). It corresponds to a cover plate or strap 
in a boiler or pipe butt joint. 

Auxiliary piece is a term used to denote any plate or other piece 
which is used in addition to the rivets and the main plates or mem¬ 
bers to form the joint. Thus, a gusset plate or a column bracket 
(Fig. 35) is an auxiliary piece. 

The term efficiency of a riveted joint is used in connection with 
boiler and pipe joints to denote the ratio of the strength of the 
joint to the strength of the solid plate. It is customary to use 
the working or allowable strengths (not allowable unit-stresses) 
rather than the maximum or ultimate strengths in calculating the 
efficiency. 

11. Modes of Failure.—A riveted joint may fail in any of the 
following ways: 

1. Shearing of the rivets as indicated in Fig. 31(a). 

2. Rupturing of the plate on a section through a line of rivets 
or on a section along a diagonal pitch as indicated in Fig. 31(6). 

3. Crushing of the rivets (or of the plate) due to the pressure 
of the plate on the side of the rivet (or of the rivet on the plate) as 
indicated in Fig. 31(c). This pressure is called the bearing pres¬ 
sure. 

4. Shearing of the plate in the margin or tearing of the plate 
in the margin, as indicated in Fig. 31(d). A marginal failure 
is usually a combination of these two actions. 

In joints having several rows of rivets the failure may be a com¬ 
bination of the above failures, as for example the rupturing of the 
plate along one row of rivets accompanied by the shearing of the 
rivets in another row. 


STRESSES IN RIVETED JOINTS 


37 


The width of the margin can be fixed arbitrarily (except as it 
affects calking of a boiler joint to secure tightness) and experi¬ 
ence has shown that if the width of the margin is 1§ to 2 times 
the diameter of the rivet the joint will not fail in the margin. 
Further, experience has shown-that failure along a diagonal will 
not occur if the transverse pitch is not less than If times the diam- 
ter of the rivet. It is evident therefore that if these conditions 
are satisfied the only stresses that need investigation in a riveted 
joint are (1) the shearing stress in the rivets, (2) the tensile stress 
in the plate on a section through a row of rivets, and (3) the bear¬ 
ing stress on the rivet (or on the plate). 

12. Stresses in Riveted Joints.—As noted in Art. 9 the equa¬ 
tion P = as is used in the analysis of the stresses in a riveted joint 



Fig. 31.—Modes of failure of riveted joints. 


in which the load on the joint is assumed to act through the cen¬ 
troid of the rivet areas, the stress on each area then being assumed 
to be uniformly distributed. 

In the equation P = as as applied to riveted joints, 

P, for a structural joint, is the force that is transmitted from 
one member of a structure to another member, and, for a boiler or 
pipe joint it is usually the force transmitted through a repeating 
group of rivets. 

The kind of stress s will be denoted by a subscript. Thus, s 3 
denotes the shearing unit-stress in the rivets; s t , the tensile unit- 
stress in the plate; and Sb the bearing unit-stress on the rivet. 
Similarly a s denotes the shearing area on which occurs, a t the 
tensile area on which s t occurs; and a& the bearing area on which s b 


occurs. 



























38 


THIN-WALLED CYLINDERS 


Other quantities used are defined as follows: 

t is the thickness of the main plates or members; 
d is the diameter of the rivets; 
p is the pitch of the rivets. 

To illustrate the method of determining the stresses in a riveted 
joint one particular joint will here be used; the method, however, 
is the same for all joints. Fig. 32(a) represents a double-riveted 
two-strap butt joint, in which it is desired to find the shearing, 




tensile and bearing unit-stresses in terms of the load P and the 
dimensions, t, d and p. 

Shearing Unit-stress .—The load P is held in equilibrium, as 
shown in Fig. 32(6), by the total resisting shearing stress, which is 
the product of the shearing unit-stress s s and the area a s on which 
the shearing stress occurs. And since each of the two rivets has 

two shearing areas, a s is equal to — T ~. Hence 


P = a s s s = 4-^-s 8 or 


s s 


= P_ 
nd 2 ' 


• (13) 
































STRESSES IN RIVETED JOINTS 


39 


It should be noted that if the above joint were a lap joint, each 
rivet would have only one shearing area, and hence the total area 

would be 2^f- instead of 4^-. 

4 4 


Tensile Unit-stress. —The total tensile resisting stress, a t s t , 
in the plate which holds the load P in equilibrium (see Fig. 32c) 
is the product of the area a t} or (p-d)t and the tensile unit-stress s t . 
Hence 


P = a t s t = {p—d)is t 


or 


P 

S ‘~ (p-d)t' 


(14) 


Bearing Unit-stress. —The total bearing stress of the plate on 
the rivets which resists the load P (Fig. 32 d) is the product of the 
bearing unit-stress s b and the bearing area a b on which the stress 
is assumed to be uniformly distributed. The area of contact be¬ 
tween the rivet and plate is a semi-cylindrical area and although 
the intensity of the pressure and the direction of the pressure prob¬ 
ably vary greatly over the area, the manner of variation is un¬ 
known. Only the component pressure parallel to the load P resists 
this load, and it is usually assumed that this component pressure 
is uniformly distributed over an area equal to the projection of 
the semi-cylindrical area of contact on a plane perpendicular to 
the direction of P, that is, an area equal to the product of the 
diameter of the rivet and the thickness of the plate. Hence, as 
indicated in Fig. 32 d, 


P = a b s b = 2id s b 


or 



. . (15) 


As noted above, the load P in boiler joints usually is taken as 
the load that is resisted by a repeating group of rivets and in 
structural joints it is the total force transmitted from one member 
to another. 


ILLUSTRATIVE PROBLEMS 

Problem 31. The members of the Fink truss shown in Fig. 33(a) are con¬ 
nected by riveted joints. The arrangement of the joint at C is shown in 
Fig. 33(6). The stress in members BC and DC are found to be 6930 lb. 
compression and 6930 lb. tension respectively. The rivets are § in. in 
diameter. Find the shearing unit-stress in the rivets connecting the 
members BC and DC to the gusset plate, and in the rivets connecting the 



40 


THIN-WALLED CYLINDERS 


gusset plate to the lower chord AC EG. Also find the bearing unit-stress of 
the members BC and DC on the rivets. 


6930 lb. 


8000 lb. 




Fig. 33.—Riveted joint in truss. 


Solution .—The shearing unit-stress in the rivets connecting members BC 
and DC to the gusset plate is 



6930 

2X0.44 


= 7870 lb. per sq. in., 


the area of cross-section of a f-in. rivet being 0.44 sq. in. 

The bearing unit-stress of the members BC and DC (or of the gusset 
plate) against the rivets is 

P 6930 

Sb = — =-= 12,300 lb. per sq. m. 

db 2X?Xs 


The shearing unit-stress in the rivets that connect the gusset plate to 
the lower chord is 


P 2X6930 cos 60° 
a s ~ 3X0.44 


6930 

F32 


= 5240 lb. per sq. in. 


The bearing unit-stress of the lower chord on the three rivets is 
P 6930 

Sb = — = -;—- = 8210 lb. per sq. in. 

ab 3X|Xf 

Problem 32 . —A boiler having a diameter, D, of 72 in. is designed to resist 
an internal steam pressure, R, of 120 lb. per sq. in. The longitudinal joint 
is a double-riveted lap joint. The thickness, t, of the plates is in., the 
diameter, d, of the rivets is f in., and the pitch, p, is 3| in. Find the 
shearing, tensile, and bearing unit-stresses in the joint. Also calculate the 
efficiency of the joint, assuming the allowable shearing, tensile, and bearing 
unit-stresses (in lb. per sq. in.) to have the following values: s s = 10,000, 
= 15,000, Sb = 22,500 (see Art. 14). 




















STRESSES IN RIVETED JOINTS 


41 


SolutiDn .—According to Art. 8 and as indicated in Fig. 34(a) the force P 
transmitted through the joint for a pitch distance p of in. along the joint, 
that is, through a repeating group of rivets (two rivets) is 


P = %RDp 

= 5X120X72X35 


= 15,120 lb. 

The shearing unit-stress in the 
rivets then, is 


P 15,120 

s s = —“ = -———- = 12,600 lb. per 
2 tt <P 2X0.60 

1 4 


sq. in. 

The tensile unit-stress in the 
plate on a section through either 
row of rivets is 


P 15,120 

( p—d)t ( 3 £ — 1)Xt6 


13,200 


lb. per sq. in. 



Fig. 34. —Stress in boiler joint. 


The bearing unit-stress of the plates against the rivets is 


Sb = 


15,120 


= 19,800 lb. per sq. in. 


Efficiency = 


2 -d-t 2X%Xi6 

least allowable strength of joint 
allowable strength of solid plate* 


ird 2 

2—X10,000 , 

4 TO 2 

Shearing efficiency e s =- , „ ^ = X f = 0.524 = 52.4 per cent. 

p-M5,000 2 p-t 

( p—d)t 15.000 p—d 

Tensile efficiency et — -— =-= 0.75 = 75 per cent. 

p-t- 15,000 p 

2 -dl- 22,500 2d 3 

Bearing efficiency =-= — X- = 0.75 = 75 per cent. 

p-M5,000 p 2 


Therefore, the efficiency of the joint is 52.4 per cent. If the joint were well 
proportioned the three efficiencies would be more nearly equal. 


PROBLEMS 

33 . A boiler 30 in. in diameter is designed to withstand a pressure of 
50 lb. per sq. in. The longitudinal joint is a single-riveted lap joint; the 























42 


THIN-WALLED CYLINDERS 


rivets have a pitch of 3 in.; the plates are f in. thick and the rivets are 8 
in. in diameter. Find the shearing, tensile, and bearing unit-stresses in 
the joint. Also find the efficiency of the joint, using the allowable unit- 
tressses in stated Prob. 32 and Art. 14. 


34. A column bracket (Fig. 35) consisting of a 6-in. by 6-in. by f-in. angle 
carries a load of 20,000 lb. and is riveted with five rivets f in. in diameter to 
a 12-in., 30-lb. channel which forms part of a 

column. The thickness of the web of the 
channel is \ in. Find the shearing unit-stress 
in the rivets and the bearing unit-stress of the 
angle on the rivets. 

Ans. s s = 66701b. per sq. in., s& = 12,200 lb. 
per sq. in. 

35 . A bridge post, Fig. 36, consists of two 
10-in., 25-lb. channels latticed together. The 
total compressive stress P in the post is 275,- 
000 lb. The post transmits this stress to a 
pin 6 in. in diameter by means of the bearing 

-in. pin- 




Fig. 35.—Column bracket. 


Fig. 36. —Column 
bearing on pin. 


plates riveted to each channel, the thickness of each web being 0.53 in. The 
rivets are f in. in diameter. Find the shearing unit-stress in the rivets and 
the bearing unit-stress on the pin. 

36. The plates of a tank 60 in. in diameter are f in. thick and are spliced 

by means of a double-riveted butt joint with two strap-plates. The strap- 
plates are § in. thick. The rivets are staggered; the two lines are 3 in. apart, 
and the pitch on each line is 3§ in. The rivets are | in. in diameter. If 
the allowable unit-stresses are: s s = 10,000 lb. per sq. in., st = 15,000 lb. per 
sq. in., and s& = 22,500 lb. per sq. in., what is the maximum internal pressure 
to which the tank can be subjected. Ans. R = 228 lb. per sq. in. 

37. A boiler 60 in. in diameter resists an internal pressure of 180 lb. per 
sq. in. The plates are f in. thick and are spliced by means of a double-riveted 
butt joint with two strap-plates. The strap-plates are f in. thick and the rivets 
are staggered; the two lines are 3 in. apart and the pitch on each line is 






























STRESSES IN RIVETED JOINTS 


43 


also 3 in. The rivets are £ in. in diameter. Find the shearing, bearing and 
tensile unit-stresses. 

Ans. s s = 6750 lb. per sq. in., Sb — 12,350 lb. persq. in., st = 10,160 lb. per sq. in. 




Fig. 37.—Riveted joint in truss. 

38 . In Fig. 37(6) is shown a joint that occurs in a riveted Pratt truss as at 
D in Fig. 37(a). If the rivets are f in. in diameter and the allowable shearing 
unit-stress is 10,000 lb. per sq. in., what maximum values can Pi and P 2 have? 
What is the bearing unit-stress on the rivets if the gusset plate and the angles, 
are f in. thick? 

39 . A triple-riveted lap joint is made of ^ in. plates. The diameter of the 

rivets is 1 in.; the pitch in the outer rows is 44 in.; the pitch in the inner row 
is 2\ in. and the distance between the rows of the rivets is 2\ iii. (a) What 
is the efficiency of this joint ? (6) If this joint is the longitudinal seam of a. 

90-in. tank, what is the maximum allowable internal pressure ? Use values 
for allowable unit-stresses stated in Prob. 32 and Art. 14. 

Ans. et = 77.8%, P = 108 lb. per sq. in. 

40. A boiler 100 in. in diameter, having plates £ in. thick, is subjected fc> 
an internal steam pressure of 160 lb. per sq. in. The longitudinal joint is a 
double-riveted butt joint with two cover plates. The diameter of the rivets 
is £ in. and the pitch is 3 in. Find the bearing, shearing, and tensile unit- 
stresses in the joint. 

13. Assumptions. Conditions Affecting Strength of Riveted 
Joints. Friction. —Since the rivets in a joint shrink while cooling, 
the plates are drawn tightly together which in turn causes friction 
between the plates when the joint is stressed. Tests 1 have shown, 
however, that slipping may occur at ordinary working stresses. 
Further, in any case, the amount of the friction is indeterminate, 
and hence the assistance obtained from the friction is usually 
neglected in the design of joints. 

Bending and Tension. —In the design of a riveted joint, the 
assumption is usually made that only shearing stresses exist on a 

1 Bulletin 49, Engineering Experiment Station, University of Illinois. 














44 


THIN-WALLED CYLINDERS 


cross-section of a rivet. The rivet, however, is always subjected 
to more or less bending, and the bending may have an important 
influence on the stresses in long rivets that connect several plates 

as indicated in Fig. 38. And, in 
a simple lap joint, bending also 
occurs as is indicated in Fig. 39. 
The effect of bending, however, 
is not disastrous if the material 
is ductile. Its influence in the 
design of the simpler types of 
joints is taken account of to 


Fig. 38.—Bending in rivet. 




Fig. 39.—Bending in rivet. 


some extent in selecting the values of the working unit- 
stresses. 

Occasionally rivets must resist a direct tensile load as in the 
arrangement shown in Fig. 40. The 
allowable tensile stress in rivets should 
always be low and the thickness of the 
head of the rivet should be investigated 
to see if the shearing area in the head is 
ample; the contraction of the rivet dur¬ 
ing cooling causes stress in the rivet of 
unknown amount, but sufficient some¬ 
times to cause the rivet head to snap 
off. Hence, rivets are not considered 
reliable for resisting direct tension. 

Rivets are cheaper than bolts, other¬ 
wise bolts would be more generally used 
since.they can be made safe in bending Fig. 40.—Rivets subjected to 
and tension as well as in shear. How- tensile stress, 

ever, with the. vibration that occurs in 

many structures it is difficult to keep the nuts of the bolts tight, 
and even the best nut-locks should be inspected frequently. 

Rivet Holes .—Rivet holes are usually formed by punching the 
plates cold, for, although holes made by drilling are preferable, 


































CONDITIONS AFFECTING STRENGTH OF RIVETED JOINTS 45 


punching is the cheaper process. There are two objections to 
punching the hole: (1) The holes seldom come into exact align¬ 
ment or register since only one plate can be punched at a time, and 
in order to cause the holes to register it is frequently necessary 
to ream out one or both holes or to use a drift pin. The latter 
method is particularly objectionable, but by either method the 
holes are enlarged and the rivets may not fill the holes, thereby 
causing excessive stress on some of the other rivets; (2) the 
tensile strength of the material around the hole is reduced due to 
the injury of the material accompanying the lateral flow of the 
metal caused by the punch. 

Both of these difficulties may greatly be reduced by making 
the punched holes somewhat under-size and then reaming them 
to the correct size, after the plates or members are out together. 
Plates are seldom punched and used in their natural punched con¬ 
dition except in the case of thin plates and in the cheaper classes 
of work. The injury done to the plate by punching may be 
partially remedied by thorough annealing; it is generally im¬ 
practicable, however, to anneal portions of large sheets merely for 
the sake of the rivet holes. Splice bars used in connecting railroad 
rails are frequently punched cold and then the whole bar annealed 
instead of being punched hot. In the best classes of marine work 
the law requires that all rivet holes shall be drilled from the solid 
plate, but the specifications for most structural work and for many 
pressure vessels require the holes to be punched small and reamed 
to the correct size. In obtaining the tensile unit-stress in a plate 
in which the holes are punched to size (not reamed) the diameter 
of the hole is sometimes assumed to be yg- inch larger than that of 
the punched hole. 

Methods of Riveting .—In structural work the riveting of some of 
the joints is done in the shop, whereas other joints must be riveted 
in the field during the erection of the structure. Shop riveting is 
usually done by machines which press the rivet in place and when 
well done is better than hand riveting. Since the rivet is a little 
smaller than the hole, the rivet while being driven, must be ex¬ 
panded throughout its whole length if it is to fill the hole, which is 
a necessary requirement for a good joint. Large rivets, therefore, 
almost always are machine riveted in order to obtain the heavy 
pressures required. 

Specifications usually require that the allowable unit-stresses 


46 


THIN-WALLED CYLINDERS 


for hand-driven field rivets shall be less (often one-third less) than 
for machine-driven shop rivets since the conditions in the field are 
difficult to control. For example, the heating of the rivets varies; 
some rivets are burned, others are underheated and may be too 
cool when driven to be made to fill the hole, etc. 

In designing joints the rivet is assumed to fill the hole and hence 
the stress, in compression members, is not increased due to having 
rivet holes in the member as is the case in tension members. 

Distribution of Load Among Rivets .—If the load on a joint is a 
central load (passes through the centroid of the group of rivets) 
the assumption is made that the load is distributed equally to all 
the rivets (see Art. 89 for a discussion of eccentric loading). 
Obviously, it is impossible for the load to be distributed in this way, 
since the load on the rivets in one row depends on the yielding of 
the rivets in the row nearer to the load and also on the yielding of 
the plate between the rows, etc. The distribution of the load, 
therefore, is not easily determined, particularly in joints having 
several rows of rivets. But, in the absence of definite information 
the assumption stated above is used, particularly in the case of the 
simpler types of joints. 

14. Allowable Stress.—The discussion, in the preceding articles, 
of the uncertainties in the actions occurring in joints should make it 
evident that the selection of the working or allowable unit-stresses 
in shear, tension, and bearing should be based not only on the 
strength of the materials of which the joint is made, but also on the 
results of tests of actual joints. Many tests on riveted joints 
have been made, and tests and experience indicate that in the 
design of joints of the usual proportions and with the methods of 
calculating stresses already discussed, the values of working 
stresses given below may be used. In using these values it is 
understood that the members or plates to be riveted meet the 
specifications for structural steel or boiler plate, etc., and that the 
rivets meet a similar requirement for rivet steel. Further, it is 
assumed that the rivet holes are punched small and reamed to 
size or are drilled from the solid plate, and that the rivets are shop 
driven. For joints having field-driven rivets or having holes 
punched to size (not reamed) the values given below should be 
reduced. 

Shearing unit-stress in rivets; s s = 10,000 lb. per sq. in. 

Tensile unit-stress in plate; st = 15,000 lb. per sq. in. 


ALLOWABLE STRESS 


47 


Bearing unit-stress on/ivets; s& = 22,500 lb. per sq. in. 

Bearing unit-stress on pins; s& = 22,500 lb. per sq. in. 

Compressive unit-stress in members; s c = 15,000 lb. per sq. in. 

Thus the ratios of the shearing and bearing allowable unit- 
stresses to the tensile allowable unit-stress are as follows: 



s&_3 
s t 2* 


These ratios are convenient for use in calculations in design. 


PROBLEMS 

41. In an ideal boiler or pipe joint the allowable strength (not allowable 
unit-stresses) in shear, bearing, and tensile are equal, (a) By equating the 
allowable shearing strength to the allowable bearing strength show that the 
diameter of the rivets in a lap joint is 2.87 times the diameter of the rivets 
(d = 2.&7t). ( b ) Similarly, show that in an ideal butt joint having two cover 

plates, d = 1.43£. Which of these two types of joints is better suited for boilers, 
pipes, etc., having thick plates ? Hint: If n represents the number of rivets 
we have for (a) 

7r ^ 2 , 4 s b 

n - =ntdsb or a=- 1. 

4 TT St 


42. By equating the allowable tensile strength to the allowable bearing 
strength show that the pitch in an ideal double-riveted lap joint is four times 
the diameter of the rivets (p = 4 d ). Also show that the efficiency of an ideal 
double-riveted lap joint is 75 per cent. 

43. Find the relation between p and d, and also the efficiency, of an ideal 
double-riveted butt joint having two cover plates. 


CHAPTER III 1 


ELEMENTARY COMBINED STRESSES AND COMBINED 
STRAINS. RESILIENCE 


15. Introduction.—The intensities of stress that are most 
easily calculated directly from the loads on a member by use of the 
equations in the preceding chapters and those immediately follow¬ 
ing may not be the most significant stresses to which the member 
is subjected. Thus, if the intensity of stress at a point on one or 
more given planes in a body is known, it may be desired to de¬ 
termine the intensity of stress at the same point but on another 
plane passing through the point. Likewise, the relations between 
the strains in various directions may be desired. A more detailed 
discussion of these topics is given in Chapter XII. 

16. Stresses on Oblique Section.—If an axial load P is applied 


1 

r 

0 

/ 

d/ 

/a' 

a 

1 

1 

[ P c«> 



(0 (P) (d) 

Fig. 41. —Stresses on oblique plane. 


to a bar (Fig. 41a), the total stress, Q (Fig. 416), on a cross-section 
perpendicular to the direction of the load is uniformly distributed 
on the area and hence is equal to the product of the area, a, of the 

1 This chapter may be omitted without causing difficulties for the student 
except, perhaps, in Chapter XII. It is strongly recommended, however, that 
Art. 16 be studied and as many more as time will permit. 

48 






















STRESSES ON OBLIQUE SECTION 


49 


section and the constant unit-stress, s. Further, since Q holds P 
in equilibrium it must be equal to, and collinear with, P and hence 
must pass through the centroid of the area a, that is, P=Q = as. 

Now, if an oblique plane is passed through the centroid, 0, of 
the area a and the lower part of the bar is removed (Fig. 41c), the 
total stress on the oblique area a' must likewise be Q in order to 
hold P in equilibrium. But Q is inclined to the area a' and, for 
convenience, will be resolved into normal (tensile) and tangential 
(shearing) components, Q n and Q s respectively (Fig. 41c), and 
since these components pass through the centroid of the area a' 
the normal and shearing intensities of stress, s n and s s on the areas 
a' are constant (Fig. 41d). Hence, the following equations^may 
be written: 

Qn = 0'S n ~ a^ri) and Qs = 0'^8 == ~ 

cos 0 .cos 0 

But 

Q n = Q cos 6 = P cos 0 and Q e =Q sin 0 = P sin 0. 

Therefore, 

P cos 0=• a s n and P sin 6= a -s s 
cos 0 cos 0 


or 


P P IP. 

s n = — cos 2 6 and s s = — sin 0 cos 0 = - — sin 20, (16) 

a a 2 a 


from which the normal and shearing unit-stresses 
on any section inclined at an angle 0 to the sec¬ 
tion on which the maximum normal stress occurs 
may be found. Now s n will be a maximum 
when 0 is zero, that is, on the plane perpendicu- 

P 

lar to P, and its value is s = — which agrees with 

the value found in Art. 3. The maximum value 
of s s in the above equation, however, occurs 
when 0 is 45° and hence, 

maximum value of s s = - —. 

z a 



Thus the maximum value of the shearing unit- Fig. 42-Stresses on 
stress in a bar subjected to an axial tensile itf teSon ^menT 
(or compressive) load is equal to one-half the ber. 
maximum normal stress developed in the bar, 
and it occurs on planes making angles of 45° with the plane on 








50 


ELEMENTARY COMBINED STRESSES 


which the maximum normal stress occurs. But, since the shear¬ 
ing strength of many materials is much less than the tensile or 
compressive strength, the shearing unit-stress developed may be 
the more significant stress. The stresses acting on the faces 
of a small block in the bar, the faces of the block being per¬ 
pendicular and parallel to the direction of P, are shown at A 
in Fig. 42, and the stresses on a block having faces making 
angles of 45 degrees with the direction of P are shown at B in 
Fig. 42. 

PROBLEMS 

44. A brick 2 in. by 4 in. by 8 in. is tested in compression by applying an 
axial load on its 2-in. by 4-in. ends. If the ultimate compressive and shearing 
strengths are 3000 lb. per sq. in. and 1000 lb. per sq. in., respectively, will the 
brick fail when subjected to a load of 20,000 lb.? 

45. A structural steel bar f in. in diameter and 10 in. long is subjected to 
an axial tensile load of 12,000 lb. (a) Find the tensile and shearing unit- 
stresses on a plane making 60° with the direction of the load. (6) Find the 
maximum shearing unit-stress. 

Ans. (a) s n = 20,300 lb. per sq. in.; $ s = ll,700 lb. per sq. in. (6) 
S s = 13,600 lb. per sq. in. 

46. Show that when a bar is subjected to an axial load the normal unit- 

1 P 

stress, s n , on the section having the maximum shearing unit-stress is-, and 

2 a 

hence has the same value as the maximum shearing unit-stress (see Fig. 42). 

47. Show that the forces (stresses) s s and s n acting on the faces (unit-areas) 
of the cube B in Fig. 42 cause an intensity of stress on the horizontal diagonal 

P 

plane through the cube equal to s or —. 

a 

17. Shearing Stresses on Planes at Right Angles. Simple 
Shear. Proposition. —If a shearing stress of intensity s s occurs on 
a plane at a given point in a body there must exist a shearing stress 
of equal intensity at that point on a plane at right angles to the 
first plane. 

Proof. —Let an elementary rectangular block (Fig. 43a) be 
removed from a body in which the block is subjected to shearing 
stresses on a pair of parallel faces, such as faces AB and CD or 
faces AC and BD. For example, the block may be in a bolt as 
shown in Fig. 43(6), where the part X causes a shearing stress of 
average intensity s h on the horizontal plane AB of part Y, and the 


SHEARING STRESSES ON PLANES AT RIGHT ANGLES 51 


part Z exerts a resisting stress of equal intensity on the face CD. 
Similarly, the part X of the shaft in Fig. 43(c) causes an average 
shearing unit-stress s v on the vertical plane BD and the part Z 
of the shaft exerts an equal resisting or opposite stress on the face 
AC. 

Now the shearing forces acting on a pair of faces form a couple 
and since the block is in equilibrium when in the body, there must 
be shearing forces on the other pair of faces such that the moments 
of the two couples are equal. If the depth of the block perpen¬ 
dicular to the paper is assumed to be unity, then 

ABs h • BD = BDs v • AB 

whence 

Sh == S v . . . .* (18) 



(a) (6) (c) 

Fig. 43.—Shearing stresses of equal intensity on planes at right angles 
to each other. 


Now if the dimensions of the block are considered to be indefi¬ 
nitely small, s h and s v may be considered to be the stress at a point. 
That is, the intensities of the shearing stresses on planes at right 
anglers to each other at any point in a body are equal; and this is 
true also when normal stresses act on the planes in addition to 
shearing stresses, as is illustrated in Fig. 42. If shearing stresses, 
only, occur on the two planes at any point in the body, the body at 
that point is said to be in a state of simple shear or of pure shear. 

18. Tensile and Compressive Stresses Resulting from Simple 
Shear. Proposition .—If a state of simple shear exists at a point 
in a body (as in Fig. 43c) there also exists normal (tensile and 
compressive) stresses on planes that bisect the planes on which the 
shearing stresses occur, and the intensities of the normal stresses 
are equal to those of the shearing stresses. 

Proof .—Let a diagonal plane AD be passed through the block 
of Fig. 44(a), or Fig. 43(a); the forces acting on one part of the 















52 


ELEMENTARY COMBINED STRESSES 


block are shown in Fig. 44(6). The face of the block in Fig. 44 is 
assumed, for convenience, to have a thickness (perpendicular to 
the plane of the paper) of unity. The force on each face is equal 
to the area of the face times the unit-stress on that face, and since 



the block ABD (Fig. 446) is in equilibrium, we have, by resolving 
the forces in the x-direction, 

ADs t =ABs s cos 45°+55s. cos 45° = s s (AB cos 45 °+BD cos 45°) 


but 
Therefore, 

Similarly, from Fig. 44(c) 


AB cos 45°-f-Z?.Dcos 45 ° = AD. 

St~ s s . 


(19) 



Fig. 45.—Brittle material when subjected to torsion fails in tension. 



Thus when a brittle material such as cast iron, which is weak 
in tension, is twisted as in Fig. 43(c) and Fig. 45(a) the material 
fails intension on a plane inclined (approximately 45°) to the planes 
on which the shearing stresses occur, as indicated in Fig. 45(6). 2 

2 The student may perform an experiment on a crayon of chalk to show the 
action indicated in Fig. 45. 
























PRINCIPAL STRESSES 


53 


19. Principal Stresses.—With any combination of stresses at 
a point in a body three planes can be found passing through the 
point, on which only normal stresses exist; the normal stresses on 
these planes, on which no shearing stresses occur, are called 
principal stresses. In many problems in the following chapters 
two of the three principal stresses are equal to zero; in some prob¬ 
lems only one of the principal stresses is equal to zero. 

Maximum normal stresses are always principal stresses, and 
hence principal stresses are of much importance in engineering 
problems; they will be discussed in greater detail in Chapter XII. 

20. Shearing Stresses Resulting from Principal Stresses. 
Proposition .—If normal (principal) stresses, only, occur on three 
planes at right angles to each other at a point in a body, shearing 




Fig. 46. —Shearing stress resulting from principal stresses. 

stresses occur on oblique planes through this point; the maximum 
value of the shearing unit-stress is one-half of the algebraic dif¬ 
ference of the maximum and minimum principal unit-stresses, that 
is, 

Max. S s = ^(Smax. s mln.) 

in which a compressive stress is to be considered a negative tensile 
stress. Further, the maximum shearing unit-stress occurs on each 
of the two planes that bisect (make 45° with) the planes on which 
the maximum and minimum normal stresses occur. 

Proof .—Let the block in Fig. 46(a) represent a small part of 
a body subjected to a tensile principal unit-stress si on the plane 
AB (and CD), a compressive principal stress S 2 on the plane BD 
(and AC), and a principal stress of zero magnitude on the faces 
parallel to the plane of the paper (bodies subjected to this state of 







54 


ELEMENTARY COMBINED STRESSES 


stress will be considered in Chap. XII.); the maximum principal 
unit-stress is +si and the minimum is — S 2 . 

Now if an oblique plane is passed through the block and a part 
of the block is removed, the forces holding the remaining part in 
equilibrium are shown in Fig. 46(6). By applying one of the equa¬ 
tions of equilibrium (resolving the forces parallel to the z-axis) 
the following equation is found; 

BC s s = AB si sin <t>-\-AC S 2 cos </>. 

Hence 

s s = [si — (— $ 2 )] sin </> cos 0 

= 2(^max.- ^min.) SU1 2 (f> .(21) 

But sin 2 (f> is a maximum (equal to unity) when 4> equals 45°, and 
hence the above proposition is shown to be true. 

It should be noted that if si and S 2 are of like sign (both 
tensile or both compressive stresses) and the third principal stress 
is zero, as in the shell of a boiler (Fig. 486), then the minimum 
stress is zero, and hence the maximum shearing unit-stress is 
-|smax. And, if two principal stresses are zero (as in Fig. 41) the 
maximum shearing unit-stress is merely Js where s is the only 
principal stress, which agrees with the value of s s found in Art. 16. 

PROBLEMS 

48. A boiler 6 ft. in diameter is made of plates f in. thick and is subjected 

to an internal steam pressure of 200 lb. per sq. in. Find the maximum shear¬ 
ing unit-stress in the plate. Ans. s s = 4800 lb. per sq. in. 

49. A gun barrel or thick cylinder (Fig. 47) on 
which hoops are shrunk (not shown in Fig. 47) is 
subjected to an internal pressure of 50,000 lb. per 
sq. in. due to the explosion of the charge; the maxi¬ 
mum radial compressive stress, s c , in the material 
is also 50,000 lb. per sq. in. Now if the maximum 
circumferential stress, s t , is 16,000 lb. per sq. in. and 
occurs at the same point in the cylinder as does s c , 
find the maximum shearing unit-stress and indicate 
the planes on which it occurs. 

50. At a certain point in a material subjected 
to stress a compressive stress of 2000 lb. per sq. in. exists in a direction 
at right angles to a tensile stress of 2000 lb. per sq. in. Find the normal 
unit-stress and the shearing unit-stress on a plane making 30° with the 
direction of the tensile stress. 

Ans. s t = 5320 lb. per sq. in. s s = 14,300 lb. per sq. in. 



Fig. 47.—Stresses in 
thick cylinder. 



STRAINS DUE TO PRINCIPAL STRESSES 


55 


21. Poisson’s Ratio and the Relation between Moduli of 
Elasticity.—If a bar is subjected to an axial tensile load the bar is 
elongated in the direction of the load, that is, in the longitudinal 
direction, and at the same time the lateral dimension of the bar 
decreases. The ratio of the lateral unit-strain to the longitudinal 
unit-strain is called Poisson’s ratio and will be denoted by the 
symbol m. The value 3 for this ratio for steel (and most structural 
metals) is approximately J; values of \ to § are frequently used. 

The shearing modulus of elasticity, E s , of a homogeneous 
material may be expressed in terms of the tensile (or compressive) 
modulus of elasticity, E, and Poisson’s ratio, m, by the following 
equation. 4 


F l 
‘ 2(1+m) 


( 22 ) 


And since both E s and E may be obtained from experimental data 
(see Art. 5, 144 and 145), this equation offers a convenient means 
of determining Poisson’s ratio. 

If the value of m for steel is taken to be J then the above expres¬ 
sion (for steel) becomes E S =%E. (Compare the ratio of the 
moduli of elasticity as given by this equation with the experimental 
values in Table 1). 

22. Strains due to Principal Stresses.—If a rectangular block 




Fig. 48.—Strain due to two principal stresses. 

of material is subjected to normal stresses, only (that is, to principal 
stresses), on two pairs of faces (Fig. 48a)—tensile unit-stresses si 

3 See “Determination of Poisson’s Ratio” by T. M. Jasper, in Proceedings 
of Am. Soc. for Testing Materials, 1924. 

4 The derivation of this equation is beyond the scope of this book; it may 
be found in treatises on the theory of elasticity. 







56 


ELEMENTARY COMBINED STRESSES 


on one pair and compressive unit-stresses S 2 on the other pair— 
the unit-elongation in the direction of si caused by Si is 

si 

ei = E’ 


and the unit-strain, e 2 , in the direction of S 2 caused by S 2 is 
S2 

€ 2 = E> 

and the unit-strain e\ in the direction of si caused by 82 is 

, S2 

ei = me2 = m _ 


in which the modulus of elasticity, E, is assumed to be the same in 
compression as in tension. Therefore, the total unit-strain in the 
direction of si is 


Sl . S 2 

t= E +m E’ 


or 


Ee = S\-\-ms2 


(23) 


And if both stresses are tensile stresses as occurs in the shell of a 
boiler (Fig. 486), and in other structures, the equation becomes 


Ee = si~ms2 


(24) 


It is important to note that unit-stress is proportional to unit- 


strain 



only when the material is subjected to one principal 


stress. Or, to put the same idea in other words, the normal 
unit-stress in a material is equal to the modulus of elasticity times 
the unit-strain in the direction of the stress (s = Ee ) only when 
the material is subjected to a normal stress in one direction. 
The use and significance of the term Ee is discussed in Art. 112 
and 113. 

23. Resilience.—Resilience is the property of a material that 
enables it to give up energy (do work) when the stress in it is 
released. The amount of energy released or recovered per unit- 
volume of the material when the intensity of stress decreases from 
the proportional limit to zero is called the modulus of resilience. 
Now the energy recovered when the stress is released from the 




RESILIENCE 


57 


proportional limit may be assumed to be equal to the work done on 
the material in stressing it to the proportional limit; that is, 
within the proportional limit, the energy lost in stressing the ma¬ 
terial may be assumed to be negligible. 5 The 
property of resilience is of special importance 
in connection with impact and energy loads 
and is discussed further in Chapter XIII and 
also in Art. 146. 

One Normal Stress .—Let a unit-volume 
(cube) of material be subjected to a normal 
unit-stress s in one direction only, as indicated 
in Fig. 49. If the stress increases gradually 

from zero value and causes the cube to elongate Fig. 49—Block sub- 
,i , , . jectedto oneprin- 

an amount e, the work done, w, is cipal stress. 

w = ise, .... (25) 

and since within the proportional limit e = ^, the work done per 

£j 

unit-volume in stressing the material to any value s less than the 
proportional limit is 

Is 2 

2 E’ . 



w 


(26) 


and hence the tensile (or compressive) modulus of resilience, k, is 


(27) 



Fig. 50.—Block subjected to pure shear. 


in which s e is the propor¬ 
tional limit of the material. 

Pure Shear .—Let a unit- 
volume of a material be sub¬ 
jected to pure shear as indi¬ 
cated in Fig. 50(a) or as 
indicated more conveniently 
in Fig. 50(5). The work 
done in gradually increasing 
the stress from zero to the 


value s s as the shearing strain increases to e, is 


5 If the stress is repeated a great many times the total energy lost may not 
be negligible although the loss of energy in any one cycle of stress may be 
negligible. 















58 


ELEMENTARY COMBINED STRESSES 


£ 

and if s s does not exceed the shearing proportional limit e 5 = -gr, 
and hence the shearing modulus of resilience, k s , is 


1 ( s s )* 2 

2 E s , 


(29) 


in which (s s )e is the shearing proportional limit. 

Combined Shearing and Normal Stresses .—If a cube of material 
is stressed as indicated in Fig. 51 the work done per unit-volume is 


_ 1 s 2 1 s 5 2 
w 2E + 2E; 


(30) 


This combination of stresses will be discussed in Chapter XII, 
and the above expression is used in Art. 113. 



Fig. 51.—Block subjected to 
tension and shear. 



Fig. 52.—Block subjected 
to two principal stresses. 


Two Principal Stresses .—If a unit-volume of material is sub¬ 
jected to normal stresses, only, on two faces (that is, to two princi¬ 
pal stresses) as shown in Fig. 52, the work done may be found as 
follows: If Si were acting alone the work done would be Jsiei, 
where ei is the deformation due to si. Now if S 2 be applied it 
does an amount of work equal to \s 2 e 2 , where €2 is the strain in 
the direction of S 2 due to S 2 . But the strain e 2 causes a strain 
equal to me 2 in the direction opposite that of si, and hence si does 
an additional amount of work equal to — Si-me 2 . Therefore the 
total work done is 

w = i siei + i s 2 e2-s 1 me2 


_ 1 si 2 1 S2 2 S1S2 
~2 E + 2 ~E m ~W’ 


(31) 


in which a tensile stress is to be taken as positive and a compressive 
strength as negative. 

















RESILIENCE 


59 


PROBLEMS 

51. Calculate the tensile modulus of resilience of structural steel using the 

values in Table 1. Ans. /s = 20.4 in.-lb. per cu. in. 

52. If the amount of energy that can be stored in a material before inelastic 
action begins is a constant regardless of the way the material is stressed, show 
that the shearing proportional limit of steel Is 0.63 times the tensile propor¬ 
tional limit, assuming E S — ^E. 


CHAPTER IV 


STRESS AND DEFORMATION DUE TO TORSIONAL 
LOADS 

24. Torsional Load, Twisting Moment and Resisting Moment 
Defined. Torsional Load. —Forces that cause a bar to twist 
about its central axis are called torsional loads. The resultant of 
torsional loads acting on a shaft is a twisting couple (two equal, 
opposite, non-coil inear forces) since simple turning or twisting can 
be produced only by a couple. The resultant moment (or moment 
of the resultant couple) which causes the twisting is the algebraic 
sum of the moments of the torsional loads about the axis of the 
bar or shaft. 

Twisting Moment. —The twisting moment for a section of a 
shaft is the name given to the algebraic sum of the moments, 
about the axis of the shaft, of the torsional loads that lie to one 
side of the section. The symbol T will be used to denote the twist¬ 
ing moment. 

It is here assumed that the torsional loads lie in planes perpen¬ 
dicular to the axis of the shaft. If the forces do not lie in such 
planes the forces may be resolved into components perpendicular 
and parallel, respectively, to the axis, the components parallel to 
the axis being neglected in the following discussion. 

Resisting Moment. —In Fig. 53(6) is shown a shaft subjected to 
a twisting moment T (equal to Pp) at one end of the shaft. Now 
since the shaft is in equilibrium the body A must exert an equal 
and opposite moment (not shown) on the other end of the shaft 
and these two moments cause the shaft to twist. 

The shaft when thus twisted has shearing stresses developed on 
each normal section of the shaft and the moment of these shearing 
stresses on any section, called the resisting moment, holds the 
twisting moment in equilibrium. For example, let a plane, BC, 
be passed through the shaft and let the portion of the shaft to the 
right of the plane be removed. Now if the remaining (left) por- 

60 


DISTRIBUTION OF STRESS ON THE CROSS-SECTION 61 


tion of the shaft (Fig. 52a) is held in equilibrium (as it was when 
in the whole shaft) a moment equal and opposite to T must be 
applied to it. This resisting moment (denoted by T r , Fig. 526) is 
exerted on the left portion by the right portion on the section DE 
cut by the plane, and is due to the shearing stresses developed on 
the section. Thus, the shearing resisting moment at a section of a 
shaft is the sum of the moments, about the center of the shaft, of 
the shearing stresses that are developed on the section in order to 



Fig. 53.—Cylindrical bar subjected to a twisting moment. 


hold the twisting moment for this section in equilibrium. Thus 
from the condition of equilibrium, we have, 

Moment of External forces = Moment of internal forces or stresses, 
that is, 

Twisting moment = Resisting moment 
or 

T= T r . 


25. Distribution of Stress on the Cross-section. —Now before 
an expression for the sum of the moments of the stresses on the 
area (resisting moment) can be found, the way in which the stress 
is distributed over the area must be determined. It is assumed 
that the intensity of shearing stress (shearing unit-stress) is zero 
at the center of the section and increases directly as the distance 
from the central axis. 

In order to supply the evidence or reasoning leading to this 
assumption it is necessary to consider: 

(a) the manner in which the shearing strains of the fibers in 
the shaft vary when the shaft is twisted. 



















62 


TORSIONAL LOADS 


(6) the relation between the shearing strain of a material and 
the accompanying shearing stress in the material. 

Information concerning these points is obtained mainly from 
the results of experiments: 

(a) Experiments have shown that when a cylindrical 1 shaft is 
twisted, a plane section of the shaft before twisting is approximately 
a plane section after twisting, and it is assumed that a diameter 
in the section before twisting is a diameter (straight line) after 
twisting. 2 If this is true, 

The shearing unit-strain of the material at any point 
in the shaft is proportional to the distance of the point from 
the central axis. 

Proof. —Fig. 54(a) represents a cylindrical shaft twisted by the 
twisting moment T or Pp. An outer element or fiber A B, having 




Fig. 54.—Shearing strain in shaft due to twisting moment. 


a cross-sectional area da, takes the form of a helix A B' when the 
shaft is twisted. Now a small portion of the fiber, which is at C 
before the shaft is twisted is at C' after the shaft has been twisted 

1 A plane section of a shaft whose cross-section is not circular does not 
remain plane when the shaft is twisted. Torsion of shafts having non-circular 
sections is discussed very briefly in Art. 31. 

2 Although this assumption seems a reasonable one, it is very difficult to 
obtain direct experimental verification. The justification for the assumption 
is to be found in the agreement of results calculated from the formula based 
on this assumption and experimental results. 











DISTRIBUTION OF STRESS ON THE CROSS-SECTION 63 

and is deformed as shown in the figure by the shearing stresses on 
its faces. 

Further, as discussed in Art. 4, the shearing unit-strain, e 3 , at 
any point of a body is equal to the tangent of the angle of deviation 
from a right angle of two planes passing through the point. 
Therefore, the shearing unit-strain, e 8 , at the surface of a cylindrical 
shaft when twisted is 

€ s = tan 0. 


But tan 0 (Fig. 54a) is equal to the arc BB' divided by the length, 
l, of the shaft, and further, BB' is equal to the radius c times the 
angle of twist, 6, expressed in radians. Therefore, 

. - BB' cd 

e s , at surface, = tan 0 = ~-j— = -j. 

Similarly, the shearing unit-strain at a point at the distance p 
from the center of the shaft (Fig. 545) is 

. ,. . DD' P 9 

€s } at distance p, = — 

Now since a radius OB before the shaft is twisted is a straight 
line, OB', after twisting, and since the lengths, l, of all fibers are 
equal, it follows that 

cd 

e s at distance c from center _ l _c 
e s at distance p from center pd p 

J 

That is, when a shaft is twisted the shearing unit-strains of the 
fibers vary directly as the distances of the fibers from the central 
axis of the shaft. 

(5) As stated in Art. 5. when a material is stressed (whether in 
tension, compression or shear), experiments have shown that for 
nearly all structural materials the unit-stress at any point in the 
material is proportional to the corresponding unit-strain at that 
point provided that the unit-stress does not exceed the propor¬ 
tional limit of the material. Hence if ( s s ) c and (e s ) c denote the 
shearing unit-stress and unit-strain, respectively, of fibers at the 



64 


TORSIONAL LOADS 


distances c from the axis, and the subscript p is used in a similar 
way, then 

= 7 ^ 4 °- = etc. = a constant. 

Me (Op 


The numerical value of this constant ratio is called the shearing 
modulus of elasticity of the material and is denoted by E s 
(Art. 5). 

Distribution of Stress over Area .—The way in which the shear¬ 
ing unit-stress varies over the area may now be stated. For, since 
the shearing unit-strains of the fibers are proportional to the 
distances of the fibers from the center of the shaft, and since the 
shearing unit-strains of the fibers are also proportional to the 
shearing unit-stresses on the fibers it follows that: 

The shearing unit-stress on a fiber of the shaft varies 
directly as the distance of the fiber from the axis of shaft 
(as shown in Fig. 55). Or, expressed mathematically, 

Os)c 0 . (^s)c ('S$)p j , , /or»\ 

7 —r— = ~ or-= -—- = etc. = a constant . . (32) 

(Ss) P P C p 


26. Expression for Resisting 



Fig. 55. —Shearing stresses in cylindrical 
shaft. 


Moment. The Torsion 
Formula. —The sum of the 
moments of the shearing 
stresses (resisting moment) 
may now be found as follows: 

The unit-stress at the 
distance p from the axis 
(Fig. 55) is M P and the 
unit-stress may be assumed 
to be constant over the ele¬ 
ment of area da. Hence, 

Total shearing stress on 
area da = (s a ) p da 

Moment of total stress on 
area da= (s 8 ) p da‘ p 


Sum of moments of stresses^on all elements of area, about axis 


of shaft 


-Tr-flft, 


dap 






EXPRESSION FOR RESISTING MOMENT 


65 


This may be written 


’,= f^dap 2 , 

J P 


(s) . (s )c 

but, as shown above,is constant and equal to ■^- 2 , and hence 

the above expression may be written 

(«.). 


T r 


J-^jdap 2 . 


Now fdap 2 is the polar moment of inertia of the area with respect 

to the central axis of the shaft. It will be denoted by the symbol 
7rd 4 

J, and is equal to (see Art. 162, Appendix II). Hence, 

1 '*H S J 

Resisting moment = T r =—, 


in which the subscript on s 8 is omitted since it will be understood 
that s s is the shearing unit-stress on the fiber at the distance c 
from the center, where c may be any value from 0 to the radius 
of the shaft, but since the maximum value of s a is vusually wanted 
the value of c will usually be the radius of the shaft. 

The Torsion Formula. —As already shown in Art. 24, the 
resisting moment holds the external or twisting moment in equi¬ 
librium and is therefore numerically equal to the twisting moment. 
Thus 

T=T r 

and hence 

s J Tc 

T = ^f or s s =±j, .(33) 

which is called the torsion formula. 

If s 5 is expressed in pounds per square inch T must be expressed 

7rd 4 

in pound-inches, c in inches and J (equal to Art. 165) in 

inches to the fourth power (in. 4 ). 

Limitations of the Torsion Formula. —The torsion formula is not 
applicable unless the following conditions are satisfied: 

1. The shaft is a circular cylinder, either solid or hollow. 

2. The loads lie in a plane or planes perpendicular to the axi& 
of the shaft. 



66 


TORSIONAL LOADS 


3. The shearing unit-stress does not exceed the shearing 
proportional limit of the material. 


ILLUSTRATIVE PROBLEM 

Problem 53. A shaft (Fig. 56) having a diameter, d, of 1.5 in. rotates at 
constant speed. It has keyed to it a driving pulley A and two driven pulleys 
B and C. The effective pull for each belt (difference between tensions on the 
two sides of a pulley) is 40 lb. per in. of width of belt. The belt on pulley A 
is 6 in. wide, and on B 4 in. wide. Pulley A has a diameter of 36 in., B of 12 
in., and C of 18 in. Find the maximum shearing unit-stress in the shaft 
between A and B ; also between B and C. 



Solution .—The twisting moment for any section between A and B is 
7 7 i = 40X6X18 = 4320 lb.-in. 

The maximum shearing unit-stress, $/, in any section between A and B is 

T d 

, TiC U 2 16 

Ss J~TrdA~ T \d* 

32 

= 4320X~ =6520 lb. per sq. in. 

7r(1.5) 3 

Similarly, the maximum shearing unit-stress, s s ", at any section between 
B and C is 

TvC c 

Ss " = ~j~ ~ (Ti—4X40X6) X 


= 3360 X—= 5070 lb. per sq. in. 





















STRESS DUE TO TORSIONAL LOADS 


67 


PROBLEMS 

54. If a twisting moment of 10,000 lb.-ft. is applied to a cylindrical shaft 

4 in. in diameter what shearing unit-stress will be developed in the outer fibers 
of the shaft? Ans. s s = 9550 lb. per sq. in. 

55. The shearing yield-point of a certain steel shaft is 30,000 lb. per. sq. in. 

and a working stress equal to one-third of the yield-point is specified. What 
should be the diameter of the shaft if the shaft is subjected to a twisting 
moment of 1200 lb.-in.? Ans. d = 0.85 in. 

56. A workman sometimes breaks a bolt in tightening the nut. If the 
workman uses a wrench with a handle which gives him a moment arm of 
15 in. and he exerts a force of 75 lb. on the handle, what should be the minimum 
diameter of a steel bolt to prevent the proportional limit of the bolt from being 
exceeded? Neglect friction. 

57. A plate A (Fig. 57) is riveted to a fixed member B by means of four 
f-in. rivets as shown in the figure. Find the shearing unit-stress at the 
center of each of the rivet areas due to the 2000-lb. loads. Is the torison for¬ 
mula applicable to this problem ? Note that the shearing area may be 
assumed to be two concentric annular areas, equivalent to the cross-sections 
of two_concentric hollow thin-walled cylinders. 




Fig. 57.—Rivets subjected to Fig. 58.—Friction clutch 

torsional shear. transmits twisting moment 

from one shaft to another. 

58. The stern end of a marine propeller-shaft has a diameter of 11 in. 

and a maximum shearing stress of 8000 lb. per sq. in. To this is connected 
a hollow steel tail shaft in which the maximum shearing stress is 10,000 lb. 
per sq. in. The internal diameter is one-half of the external diameter. Find 
the diameters. Ans. ch = 5.21 in., d 2 = 10.4 in. 

59. A force P of 196 lb. (Fig. 58) will cause a normal pressure of 15 lb. per 
sq. in. on the rubbing surfaces of the friction clutch shown. If the coefficient 









































68 


TORSIONAL LOADS 


of friction is 0.25 and the diameter of the shaft is 2 in., find the maximum 
shearing unit-stress in the shaft. 

60. In the band brake shown in Fig. 59 the force P is 100 lb., the angle of 
contact, a, is 270°, and the coefficient of friction n, is 0.2. Therefore, the 



Fig. 59.—Band brake transmits twisting moment to shaft. 


tension in the band at C is T c = 1300 lb. and, if the maximum friction is being 
developed, the tension in the band at B is Tb^TaXc^ =3330 lb. (a) If 
the diameter of the shaft is 3 in. what is the maximum shearing unit-stress in 
the shaft? Ans. s s = 3830 lb. per sq. in. 



5 


• 18 - 


H - 36 -- >J 




Fig. 60.—Twisting moment ex- Fig. 61.—Screw of press subjected to 
erted on shaft by means of torsion due to pull on handle, 

friction drive. 


61. Compare the strengths of the following shafts when stressed to their 
limits of proportionality. The first is solid 21 in. in diameter and the shearing 
proportional limit is 25,000 lb. per sq. in.; the second is hollow, having an 


































































TWISTING MOMENT IN TEEMS OF HORSEPOWER AND SPEED 69 


outside diameter of 18 in. and an inside diameter of 9 in., and a proportional 
limit of 45,000 lb. per sq. in. 

62. A crown friction drive as shown in Fig. 60 is used on screw-power 

presses, motor trucks, etc. The cast-iron disk B rotates at 1000 r.p.m. and 
drives the crown wheel C, which in turn drives the chain sprocket E. The disk 
C is faced with leather-fiber for which the coefficient of friction is 0.35. If 
the diameter of the disk C is 20 in. and the normal pressure of B against C is 
343 lb. what should be the diameter of the shaft between C and E in order to 
prevent the shearing unit-stress due to the twisting moment from exceeding 
8000 lb. per sq. in. (The working stress is taken low because the load is not 
a steady or static load.) Ans. d = 0.915 in. 

63. The shaft-straightening hand press shown in Fig. 61 is used for bending 
or straightening steel shafts requiring a pressure, Q, of about 24,000 lb. The 
operating screw has a root diameter of 2 in. and there are four square threads 
per inch. If the coefficient of friction is 0.2, a force of 161 lb. applied on the 
handle at A will cause the required force Q on the shaft. Calculate the maxi¬ 
mum shearing unit-stress in the shaft. 


27. Twisting Moment in Terms of Horsepower and Speed.— 

Frequently the forces acting on the shaft are unknown, and the 
twisting moment must be found from the 
horsepower transmitted by the shaft and the 
speed (angular velocity) of the shaft. This 
may be done as follows: Let a pulley A (Fig. 
62) be keyed to a shaft B and transmit a 
twisting moment T (equal to Pp) to the shaft 
when rotating at a constant angular velocity 
of co radians per second or n revolutions per 
minute. The work done by the twisting 
couple T, in any angular displacement is the product of the 
moment of the couple and the angular displacement (in radians). 
Hence, _j j 

Work done on shaft per second = Tco 

rr2irn . e l'wn .. 

= T-^7t, since co = -7T7t radians. 
60 60 



If the twisting moment T is expressed in inch-pounds the work 
done will also be expressed in inch-pounds. Now the rate of doing 
work at 550 ft.-lb. per sec. or 550X12 in.-lb. per sec. is defined 
to be a horsepower, and hence the number of horsepowers (h.p.) 
transmitted by the shaft is 

fb ) — T27rn 

60x550X12 








70 


TORSIONAL LOADS 


in which T is expressed in lb.-in. and n in r.p.m. Hence 


y_ 60X550X12X(h.p. ) 


2im 


which expresses the relation between twisting moment, horse¬ 
power, and speed. 


PROBLEMS 


64. The hydraulic turbines in the Keokuk water-power plant are rated at 
10,000 h.p., with an overload capacity of 13,500 h.p. The vertical shaft con¬ 
necting the turbine and the generator is 25 in. in diameter and rotates at 
57.7 r.p.m. What is the maximum shearing unit-stress developed in the shaft 
when developing full load and when developing maximum overload? 

65. A pulley 10 ft. in diameter is mounted on a 2^-in. shaft and rotates at 

150 r.p.m. transmitting power to a belt. If the belt tensions are 1800 lb. and 
1400 lb. (a) what horse-power is transmitted ? (6) What is the maximum 

fiber stress in the shaft ? Ans. (a) (h.p.) =114, ( b) s s = 5220 lb. per sq. in. 

J 66. A hollow shaft is used to transmit 6000 h.p. The outside diameter 
is 18 in., the speed is 90 r.p.m., and the maximum shearing unit-stress devel¬ 
oped in the shaft is 12,000 lb. per sq. in. Find the inside diameter of the shaft. 


Ans. di = 16.5 in. 


67. What horsepower can a solid steel shaft 6 in. in diameter transmit when 
rotating at 120 r.p.m. and developing a shearing unit-stress of 10,000 lb. per 
sq. in. in the outer fiber? 

28. Angle of Twist of Cylindrical Shaft.—In some problems 
the size of shaft required to transmit a given twisting moment (or 
given horsepower at a given speed) is controlled by the allowable 
angle of twist rather than the allowable shearing unit-stress. In 
other words, stiffness rather than strength is the controlling factor 
in the design. The angle of twist, 0, (Fig. 54) caused by a given 
twisting moment T may be found as'follows: 

As explained in Art. 5, 


— = a constant = E , 




provided that s s does not exceed the shearing proportional limit of 
the material. But as shown in Arts. 25 and 26 


Tc 
Ss = -r 


J 




ANGLE OF TWIST OF CYLINDRICAL SHAFT 


71 


Therefore, 

Tc 

F =L = T1 
hs Cd Jd 

l 

Hence, the angle of twist, 6, when not greater than that corre¬ 
sponding to the proportional limit of the material is, 

Tl 

e= EJ’ . (35) 

in which 6 is expressed in radians, any consistent units of force 
and length being used to express the other quantities. Thus, T 
is usualfy expressed in inch-pounds, l in inches, E s in pounds per 
square inch, and J in inches. 4 


PROBLEMS 

68. A steel shaft 4 in. in diameter transmits 200 h.p. at a speed of 250 
r.p.m. The distance between the driving and driven pulleys is 10 ft. De¬ 
termine whether the following two requirements are satisfied: 

(а) Maximum shearing unit-stress not to exceed 10,000 lb. per sq. in. 

(б) Twist of shaft not to exceed 1 degree per 20 diameters of length. 

Ans. s s = 4020 lb. per sq. in. 0 = 0.77° per 20 diam. 

69. A wrought-iron shaft 7.5 ft. long and 2 in. in diameter twists through 
an angle of 10.5 degrees when resisting a twisting moment of 2500 lb.-ft. 
The shearing proportional limit of the material is 24,000 lb. per sq. in. Find 
the value of the shearing modulus of elasticity. 

Ans. E s = 9,360,000 lb. per sq. in. 

70. A hollow bronze shaft having an outside diameter of 6 in. and an 
inside diameter of 4 in. is 12 ft. long. If the shaft is twisted by applying 
moments at its ends, what will be the angle of twist when the maximum shear¬ 
ing unit-stress is 8000 lb. per sq. in. (E s = 6,000,000 lb. per sq. in.) 

71. The steel shaft connecting a hydraulic turbine with an electric generator 
was so long that it twisted enough to cause electrical trouble in the generator. 
If the shaft had been replaced by one with the same diameter but made of 
steel having a greater shearing strength would the amount of twist have been 
decreased? 

29. Shaft Couplings. —Two shafts frequently are connected by 
means of a bolted coupling (Fig. 63a) so that, as the twisting 
moment is transmitted from one shaft to the other, shearing stresses 
are developed in the bolts. For example, let the shaft A (Fig. 63a) 



72 


TORSIONAL LOADS 


exert a twisting moment on the shaft B just sufficient to overcome 
the resistance to motion of shaft B, the resisting moment exerted 
by B being equal to the moment of the shearing stresses in the 
bolts as indicated in Fig. 636. 



(a) (6) 

Fig. 63.—Shear in bolts of coupling. 


Now if the diameter, d, of the bolts is small in comparison with 
the distance, r, from the center of the bolts to the center of the 
shaft, the shearing unit-stress, s„ over the cross-section area, a, 
of each bolt may be assumed to be constant. Hence the resisting 
moment, T r , is 

T r =m(as s -r) 

ird 2 

= m—s s • r, 

in which m is the number of bolts, all of which are on a circle of 
radius r. And since this resisting moment is equal and opposite 
to the twisting moment T, then 

T= m 7 ~s s ^r .(36) 

If T is expressed in inch-pounds, d'and r must be expressed in 
inches, and in pounds per square inch. But from Art. 27 

^ _ 60 X550 X 12 X (h.p.) . „ 

-t---~-m.-lb. 

27rn 

Therefore, 

60 X 550 X12 X (h.p.) xd 2 

- 2^ - -=m—s,-r . . . (37) 

from which any one of the quantities may be found if all the others 
are known. 













STRESS BEYOND PROPORTIONAL LIMIT 


73 


The resisting moment, T r , could have been obtained from the 
expression ~~ in which 

J = m(J+ar 2 ), see Art. 164 


but J = xd, 4 and since d is small in comparison with r then d 2 
is negligible in comparison with r 2 . Hence the resisting moment 
is 


T r 


s s mar 2 ird 2 

■ - = m—r-rs & 

r 4 


which is the same expression as found by the method used above. 


PROBLEMS 

72. A solid shaft 4 in. in diameter is to be connected to another shaft of 
the same size by means of a coupling as shown in Fig. 63a. If six f-in. bolts 
are used on a circle having a diameter of 10 in., what (h.p.) can the shaft trans¬ 
mit when rotating at 150 r.p.m. and when the shearing unit-stress in the bolts 
is 10,000 lb. per sq. in. What will be the maximum shearing unit-stress in 
the shaft? 

73. A coupling is to be used to connect two shafts having diameters of 

4 in. The maximum allowable shearing unit-stress in the shafts is 10,000 lb. 
per sq. in., the diameter of the bolt circle is 8 in. and the allowable shearing 
unit-stress in the bolts is 9000 lb. per sq. in. Find the number of f-in. bolts 
necessary. Ans. Eight. 

30. Stress Beyond Proportional Limit. Modulus of Rup¬ 
ture.—As shown in Art. 26 the value of s s in the torsion formula, 



of the shaft only when the proportional limit of the material is not 
exceeded. If the twisting moment, T, causes a unit-stress greater 
than the shearing proportional limit of the material the internal or 
resisting moment is still equal to (holes in equilibrum) the external 
or twisting moment, but the resisting moment is not given by the 
s J 

expression since this expression is found by assuming that the 
c 

shearing unit-stress varies directly as the distance from the center 
of the shaft which is true only when the stress does not exceed the 
proportional limit of the material. 

This fact is illustrated in Fig. 64. A diameter before twisting 
is assumed to be a diameter (straight line) after twisting even 



74 


TORSIONAL LOADS 


though the material is stressed beyond the proportional limit, and 
hence the unit-strain in each of the two shafts (Fig. 64), varies 
directly as the distance from the center of the shaft. But in the 
shaft that is stressed beyond the proportional limit (Fig. 646) the 
unit-stress is not proportional to the unit-strain except near the 
center of the shaft where the strain is small, and hence the unit- 
stress does not vary directly as the distance from the center of 
the shaft except for a short distance out from the center. If the 
material of the shaft is ductile enough to have a yield-point the 
unit-stress will vary about as shown in Fig. 64(6) when the yield- 
point of the shaft is reached. 



Fig. 64.—Distribution of shearing stress in shaft when stressed 
above proportional limit. 

Modulus of Rupture .—Now the moment of the stresses in Fig. 
64(6) is larger, for a given unit-stress at the surface of the shaft, 
than it would be if the unit-stress varied directly from zero at the 
center to this given value at the surface as indicated by the dotted 
line. Therefore, if the shearing ultimate strength 3 of the material 



found for T would be smaller than the maximum twisting moment 
that can be applied to the shaft. Tests show that the maximum 
twisting moment that can be applied to cylindrical steel bars is from 
10 to 20 per cent greater than that calculated from the torsion 
formula by substituting the shearing ultimate strength for s s . 

The value of s a found from the torsion formula by substituting 

8 The shearing yield-point of a ductile material is regarded as its useable 
shearing ultimate strength. 














TORSION OF NON-CIRCULAR SECTIONS 


75 


for T the value of the maximum twisting moment that a shaft 
resists when tested to rupture is called the shearing modulus of 
rupture and will be denoted by s r . Thus 

„ _ Tm ax . C 

Sr ~ J * 


It should be noted that s r is not the unit-stress in the material 
caused by the moment T max ., and, as already noted, it is not equal 
to the shearing ultimate strength 4 of the material; it is merely a 
value (expressed in lb. per sq. in.) from which the maximum twist¬ 
ing moment that a cylindrical shaft can resist, may be found. 
31. Torsion of Non-circular Sections.—As stated in Art. 26 

Ss j 

the torsion formula, T=—, applies only to a shaft having a cir- 
c 




Fig. 65.—Stress in shafts having non-circular sections. 


cular cross-section, in which case a plane section of the shaft before 
twisting is a plane section after twisting and the shearing unit- 
stress varies directly as the distance from the center, the unit- 
stress being maximum at the outer fiber. 

If a bar has an elliptical, rectangular, or other non-circular 
cross-section, a plane section before twisting becomes a warped 
surface after twisting, and, if the section is free from re-entrant 
angles, the maximum unit-stress occurs on the surface fiber nearest 
to the center of the shaft instead of the fiber most remote from the 
center. The distributions of stress on the cross-sectional areas of 
elliptical and rectangular bars are shown in Fig. 65. 

The derivation of the equations giving the maximum shearing 
unit-stress and angle of twist for bars that have non-circular cross- 
sections is beyond this scope of this book. 

4 For a method of finding the shearing ultimate strength of a material from 
a torsion test of a solid cylindrical specimen see Upton’s Materials of Construe- 
tion_(John Wiley & Sons), p. 52. 













76 


TORSIONAL LOADS 


Torsion of Members Other than Shafts .—Frequently structural 
members such as I-beams, flat plates, columns, etc., in buildings, 
ships, locomotives, cranes, etc., are subjected to twisting actions, 
and although the twisting in such members is, in general, considered 
to produce secondary effects, the twisting may have a marked in¬ 
fluence on the stiffness and strength of the member, particularly 
since members that are designed to resist bending have, in general, 
relatively small resistance to torsion; torsional action in structural 
members should therefore be reduced to a minimum, or allowance 
should be made for the effects of the torsion. 


CHAPTER V 


TRANSVERSE LOADS. STRESSES IN STATICALLY 
DETERMINATE BEAMS 

32. Preliminary Considerations.—A beam is a bar that is bent 
by forces acting perpendicular to the axis of the bar. 1 Forces that 
act on a beam are called transverse or cross-bending loads. It will 
be assumed that the cross-bending loads acting on a beam lie in 
one plane containing the central longitudinal axis of the beam and 
that this plane is a plane of symmetry of the beam. 

Beams are important members in many engineering structures 
and machines. Several common types of beams are described as 
follows: 

A simple beam is one that rests on two supports at the ends of 
the beam. Fig. 66 shows a horizontal simple beam subjected to 
two equal concentrated loads at the third-points. 

P P 


Fig. 66.—Simple beam. Fig. 67.—Fixed beam. 

Affixed or fixed-ended beam is one so restrained at its ends that 
the slope of the curve of the beam at the restrained ends does not 
change when the load is applied. Fig. 67 shows a beam fixed at 
both ends, and subjected to a uniformly distributed load of w 
pounds per foot; Fig. 68 shows a beam fixed at one end and sup¬ 
ported at the other, and subjected to a concentrated load at mid¬ 
span. The end connections of beams in structures and machines 

1 If the beam acts as it is assumed to act, in this and the following chapters, 
its length must be at least several times its depth. Further, the thickness or 
width of the beam must be sufficient to prevent collapse by wrinkling; thus, 
extremely deep and extremely thin beams are excluded. 

77 












78 


TRANSVERSE LOADS 


frequently offer considerable restraint, but not enough that the 
beams may be considered “ fixed” ; such beams are then inter¬ 
mediate between simple beams and fixed beams. 


P 



Fig. 68. —Beam fixed at one end, Fig. 69. —Cantilever beam, 

supported at other end. 


A cantilever beam is one that is fixed at one end and free at the 
other. Fig. 69 shows a cantilever beam subjected to a uniformly 
distributed load over its entire length, and to a concentrated load 
at its end. 

Fig. 70 shows an overhanging beam that overhangs both sup¬ 
ports and that carries concentrated loads at the ends of the beam 
and a uniformly distributed load over the span between supports. 



Fig. 70.—Overhanging beam. Fig. 71.—Continuous beam. 


A continuous beam is one that rests on more than two supports. 
Fig. 71 shows a continuous beam with three equal spans carrying 
a uniformly distributed load of w\ pounds per foot over two spans 
and a uniformly distributed load of wo pounds per foot over the 
third span. 

Pure bending is bending caused by couples. Thus the mid- 
third of the beam in Fig. 66 is subjected to pure bending. Bending 
produced by forces that do not form couples is called ordinary 
bending. As will be shown later, a beam subjected to pure bend¬ 
ing has no shearing stresses developed in it; the stresses on any 
section are normal (tensile or compressive) stresses whereas in 
ordinary bending the loads develop shearing stresses as well as 
normal stresses. 

Reactions. —For convenience the forces exerted on a beam by 
the supports are called reactions and the other forces are called 























PRELIMINARY CONSIDERATIONS 


79 


loads, but both loads and reactions are merely external forces that 
act on the beam and hold the beam in equilibrium. 

If the loads are known, the reactions can be found by use of the 
equations of equilibrium provided that there are not more than two 
reactions . For, the forces that hold the beam in equilibrium con¬ 
stitute a parallel force system in a plane, and for such a force 
system there are only two independent equations of equilibrium; 
namely, 

2F = 0 2M a = 0 

or 

XM = 0 ZM b = 0 

which state that the algebraic sum of the forces and the algebraic 
sum of the moments of the forces about any point are equal to zero. 
Or, the algebraic sum of the moments of the forces about each of 
two points A or B, not in a line parallel to the forces, are equal to 
zero. 

Statically Determinate and Statically Indeterminate Beams .— 
Beams for which the reactions can be found from the equations of 
equilibrium are called statically determinate beams, and those for 
which the number of unknown reactions is greater than the 
number of equilibrium equations are called statically indeterminate 
beams . Simple and cantilever beams, and overhanging beams 
that rest on two supports, are statically determinate beams, whereas, 
fixed-ended beams and continuous beams are statically indetermin¬ 
ate, and hence require equations in addition to the equilibrium 
equations in order to determine the reactions. 

This chapter deals mainly with the problem of determining the 
relation between the external forces acting on statically determinate 
beams and the stresses which the forces develop in the beams. 
Statically indeterminate beams are discussed in Chapters VII and 
IX. 

PROBLEMS 


74. Find the reactions of the supports for the beam shown in Fig. 72; 
neglect the weight of the beam. 



20C0 lb. 

1000 lb. 



500 lb. per ft. 

400 lb. per ft. 




'IIIIIIIIIIIIIIIIII111U, 

11111111! 1! 1111 LLLb 

-7S- 

'II1111111111 



k-4^4 


12 - 


|<—4-—>|<-6' 


^<— 3 - 


Fig. 72. 


Fig. 73. 















80 


TRANSVERSE LOADS 


75. Find the reactions of the supports for the beam shown in Fig. 73; 
neglect the weight of the beam. Ans. 7^ = 4267 lb. 72 2 = 5133 1b. 


33. Vertical Shear, Resisting Shear, Bending Moment, and 
Resisting Moment.—In Fig. 74(a) is shown a simple beam sub¬ 
jected to two concentrated loads, the weight of the beam being 
negligible. The reactions Ri and R 2 are found to be 6500 lb. and 
3500 lb., respectively. Let it be required to determine the 
character of the internal forces (stresses) that must occur on any 
section of the beam, such as section A A. If the portion of the 
beam to the right of section A A is removed, forces equivalent to 
those that the right portion exerted on the left portion must be 
applied on section A A in order to hold the left portion in equili¬ 
brium, since it was in equilibrium before the right portion was 




O) 


Fig. 74. —Stresses at any section hold external forces in equilibrium. 


removed. The problem, therefore, is to determine the forces 
(stresses) that must act on section A A in order to hold the forces 
Ri and P\ in equilibrium. 

Now, it is only the unbalanced part, or resultant, of the external 
forces that must be held in equilibrium by the stresses on the sec¬ 
tion of the beam, and hence, it is convenient to have a name for 
the resultant, or stress-producing part, of the loads. Thus, 

The vertical shear for a section of a beam is the magnitude 
of the resultant of the forces (loads and reactions) that lie 
on one side of the section. Or, in other words, it is the 

algebraic sum of the forces that lie on one side of the section. 

¥ 

The symbol V will be used to denote vertical shear and, for 
convenience, the forces that lie to the left of the section will, as a 
rule, be used. 

The total shearing stress on a section of the beam is called the 











BENDING AND RESISTING MOMENTS 


81 


resisting shear at the section and will be denoted by the symbol 
V r . And since, from the condition of equilibrium, the algebraic 
sum of all the forces acting on the left (or right) portion of the beam 
must equal zero we have, 

Vertical shear = Resisting shear 

or, 

V= V r . 

For the beam shown in Fig. 74a the vertical shear V for section 
A A is a downward force equal to 8000 lb.— 6500 lb. or 1500 lb., 
and hence the resisting shear, V r , on section A A (Fig. 746) is an 
upward stress of 1500 lb. 

Bending Moment and Resisting Moment. —The left portion of 
the beam (Fig. 746), however, will not be in equilibrium unless the 
algebraic sum of the moments of the forces acting on it are also 
equal to zero. Now the moment of Pi about a line in the section 
A A is greater than the moment of Pi, and hence this unbalanced 
moment of the external forces must be resisted by a moment 
exerted by the forces acting on the section A A. These forces on 
the section A A are the normal stresses exerted by the right portion 
of the beam on the left portion and consist of compressive stresses 
on the upper part of the section and tensile stresses on the lower 
part. 

Since only the unbalanced part of the moments of the external 
forces is effective in producing tensile and compressive stresses in 
the beam, it is convenient to have a name for this stress-producing 
moment. Thus, 

The bending moment at a section of a beam is the moment, 
about the section, of the resultant of the external forces 
that lie on one side of the section; that is the algebraic sum 
of the moments, about the section, of the external forces 
that lie on one side of the section. 

The bending moment will be denoted by the symbol M and the 
forces that lie to the left of the section will, as a rule, be used. The 
bending moment at section A A (five feet from Pi) is, 

M = 6500X5-8000X2 =16,500 lb.-ft. 

Now, as noted above, the unbalanced moment of the external 


82 


TRANSVERSE LOADS 


forces (bending moment) about the section A A is resisted by the 
moment of the tensile and compressive stresses on the section. 
This moment is called the resisting moment . Hence, 

The resisting moment at a section of a beam is defined 
to be the algebraic sum of the moments of the stresses 
acting on the section, about a line in the section. It will 
be denoted by the symbol M r . 

But from the condition of equilibrium the sum of the moments 
of all the forces acting on the left portion of the beam (Fig. 746) 
must equal zero. Therefore, 

Moment of external forces = Moment of internal forces; 
that is, 

Bending moment = Resisting moment 
or 

M = Mr. 


Further, from the condition of equilibrium, the sum of all the 
horizontal forces acting on the left portion of the beam (Fig. 746) 
must equal zero. And since no horizontal external forces act on 
the beam, we have 

Algebraic sum of the horizontal stresses = 0. 


That is, the sum of the compressive stresses, denoted by C (Fig. 
746), equals the sum of the tensile stresses, denoted by T. Hence, 
T=C. 

Summarizing: Since any portion of the beam (Fig. 746) is in 
equilibrium the forces acting on the portion must satisfy the 
following equations: 


Algebraic sum of horizontal stresses = 0, 
Algebraic sum of vertical forces = 0, 
Algebraic sum of moments of forces = 0, 


or T=C 
or V=V r 
or M=M r .( 38) 


An expression will now be found for the resisting moment M r 
in terms of the tensile or compressive unit-stress at any point in 
the cross-section of the beam, and the dimensions of the cross- 


EXPRESSION FOR RESISTING MOMENT 


83 


section. The shearing unit-stress in the beam will be discussed 
in Art. 40. 

In discussing the stresses in a beam it is convenient to con¬ 
sider the beam to be composed of fibers, a fiber being a rod having 
an elementary cross-sectional area da and extending the length of 
the beam parallel to the axis of the beam; a fiber unit-stress then is 
the intensity of stress on a section of any fiber, and in general may 
be either a normal or a shearing unit-stress although for the pres¬ 
ent the normal unit-stress only will be considered. 

34. Expression for Resisting Moment. The Flexure 
Formula.—In order to find the sum of the moments of the stresses 
on the fibers at any section of the beam, that is, in order to find 
the resisting moment at any section, the way in which the fiber 
unit-stress varies with the position of the fiber in the beam must 
be known, that is, the law of distribution of the intensity of stress 
over the section must be known. 

Distribution of Stress on Section .—In order to show how the 
unit-stress on a fiber varies with the position of the fiber in the 
beam there is required a knowledge of: 

(а) The way in which the strain of a fiber varies with the posi¬ 
tion of the fiber in the beam when the beam is bent, and 

(б) The relation, for the material of which the beam is made, 
of the unit-stress in a fiber to the strain of the fiber. 

Information concerning these points comes mainly from the 
results of experiments: 

(a) When a simple horizontal beam is bent the fibers on the 
top side shorten and those on the bottom elongate, and the fibers 
in one plane within the beam do 
not deform. This plane is called 
the neutral surface, and the line of 
intersection of the neutral surface 
and a cross-section of the beam is 
called the neutral axis for the 
section. Further, when a beam 
is subjected to bending, experi¬ 
ments show that if two straight lines, DE and FG (Fig. 75) are 
drawn, before the beam is bent, on the side of the beam a short 
distance, MN, apart, these lines will still be approximately 





84 


TRANSVERSE LOADS 


straight lines, D'E' and F'G', after the beam is bent. 2 Thus, one 
fiber M N, in the neutral surface, remains constant in length and 

The strain of any fiber is directly proportional to the 
distance of the fiber from the neutral surface. Further, 
since the original lengths of all the fibers are equal 3 the 
unit-strain of any fiber is also proportional to the distance 
of the fiber from the neutral surface. 


(6) Now if one of the fibers were removed from the beam and 
subjected to an axial load in a testing machine so that the unit- 

stress ^equal to ^ and the unit-strain ^equal to could be 

measured it would be found, as stated in Art. 5, that the unit- 
stress on the fiber is proportional to the unit-strain of the fiber 
'provided that the proportional limit of the material is not exceeded. 
Further, it is assumed that the fiber when in the beam acts accord¬ 
ing to the same law as when tested alone. 

Therefore, since the unit-strain of a fiber is directly propor¬ 
tional to the distance of the fiber from the neutral surface, and 
since the unti-stress on a fiber is directly proportional to the unit- 
strain of the fiber, it follows that 


The unit-stress on a fiber of a beam at any section of the 
beam is directly proportional to the distance of the fiber 
from the neutral axis. 


2 Since the straight lines may be considered to be traces of cross-sectional 
planes, this fact is sometimes stated: “A plane section before bending is a 

plane section after bending;” the essential fact, how¬ 
ever, is not that a plane section is conserved, but 
that a fiber changes its length and that this longitu¬ 
dinal strain is proportional to the distance of the fiber 
from the neutral surface. The selection of plane 
sections is mainly a matter of convenience in explana¬ 
tion. If shearing stresses exist in the beam, two par¬ 
allel plane sections DE and FG (Fig. 76) before bending 
become curved surfaces D'E' and F'G' after bending 
but the longitudinal strains of the fibers are affected very little by the shear¬ 
ing deformation in beams of the usual proportions. 

3 If the beam is a curved beam (not straight before it is loaded) the lengths 
of the fibers between two normal sections of the beam would not be equal and 
hence, although the total deformation would be proportional to the distance 
from the neutral axis, the unit-strain would not be proportional to this distance. 
This demonstration is, then, limited to straight beams. 




EXPRESSION FOR RESISTING MOMENT 


85 


Thus, if Sy and s c (Fig. 77) denote the unit-stresses on fibers at 
the distances y and c, respectively, from the neutral axis the above 
statement is expressed mathematically as follows: 

— = - or — = — = a constant. 
s c c y c 

Expression for Resisting Moment .—An expression may now be 
found for the resisting moment (the algebraic sum of the moments 
of the stresses on the cross-section about the neutral axis) in terms 
of the unit-stress on any fiber and the dimensions of the cross-sec¬ 
tion of the beam as follows: 



Fig. 77. —Intensity of stress varies directly as distance from neutral axis. 


The unit-stress, s v , on a fiber at the distance y from the neutral 
axis (Fig. 77) may be assumed to be constant over the cross- 
sectional area, da, of the fiber, and hence, 

Total stress on area of one fiber = s y da 
Moment of total stress on one fiber = s v yda 

Sum of moments of stresses on all fibers = M r = j" s y y da. 

This may be written 

S yV 2 da, 



s s 

and since, as shown above, — = — = a constant, this may be written 

V c 


M 


= S yJ' ° r ^ r= Hj'd 2( ^ a ■ 


But the expression Jy 2 da is the moment of inertia of the 


cross-section of the beam with respect to the neutral axis, since y 
is measured from the neutral axis (see Appendix II.). Therefore, 

si 

Resisting Moment = M r = —, 

c 










86 


TRANSVERSE LOADS 


in which s is the tensile or compressive unit-stress on a fiber at 
the distance c from the neutral axis, and since the maximum value 
of s is usually desired, the distance c will usually be taken as the 
distance to the most remote fiber. 

Flexure Formula .—But as shown in Art. 33, M = M r . 

Therefore, 

M=— .(391 

c 


which is called the flexure formula. It expresses the relation 
between the external forces acting on the beam, the tensile or com¬ 
pressive unit-stress at any point in the beam, and the dimensions 
of the cross-section of the beam. If s is expressed in pounds per 
square inch as is customary in the United States then M must be 
expressed in pound-inches, I in inches 4 , and c in inches. Further, 

- is called the section modulus of the beam and is expressed in 
c 

inches 3 . 

Position of the Neutral Axis .—The value of I in the flexure 
formula cannot be found unless the position of the neutral axis in 
the area is known. Now, as noted in Art. 33, the sum of the com¬ 
pressive stresses on the section must be equal to the sum of the 
tensile stress on the section, that is, the sum of all the horizontal 
stresses at any section must be equal to zero; this condition serves 
to locate the neutral axis. Thus, 


S horizontal stresses on section = 0, 


-yda= 0, 

y 


that is (see Fig. 77), 
or 

and hence 

But ~ is not equal to zero. Therefore, 


^ Syda = 0, 

f 


~^Vda =0 


, since —=a const. 

y 


Jyda = 0, 

in which y is measured from the neutral axis, but Jyda is, by 



THE FLEXURE FORMULA 


87 


definition, the moment of the cross-sectional area of the beam with 
respect to the neutral axis (see Fig. 78), and may be written ay in 
which a is the total area of the cross-section and y is the distance 
of the centroidal axis from the neutral axis, about which moments 
are taken. Thus, 


yda = ay — 0. 


But a is not equal to zero. Therefore, 

y = 0, Fig. 78. 

which states that the distance from the centroidal axis to the 
neutral axis is zero, and hence the neutral axis is coincident with 
the centroidal axis. It is assumed that the student is familiar 
with the methods of locating the centroids of area; this topic, 
however, is treated in Appendix I. Further, the methods for 
determining the moment of inertia of an area with respect to an 
axis is discussed in Appendix II. 




5^ \ Assumed position 

^ \\, of neutral axis 

"* > \ 

^Centroidal 
axis 

da 


ILLUSTRATIVE PROBLEMS 

Problem 76. A simple beam having a rectangular cross-section (Fig. 79) 
is subjected to a uniformly distributed load of 400 lb. per ft. (including the 
weight of the beam) over the whole span, and a concentrated load of 2000 lb. 
at a distance of 4 ft. from the left support. Find the tensile unit-stress on 
the outer fiber of the beam at the section AB, 5 ft. from the left support. 




Solution.—First Method .—The reactions are found to be 
R, = 47001b., #2 = 3700 lb. 

The bending moment M at the section AB is 

M = 4700X5-400X5X2.5-2000X1 
= 16,500 lb.-ft. = 198,000 lb.-in. 






















































88 


TRANSVERSE LOADS 


The centroidal axis (and hence the neutral axis) of the cross-section is the 
central horizontal axis, xx, and the moment of inertia of the cross-section 
about the neutral axis is (see Art. 165) 

/= T \bd 3 = rs6(12) 8 = 864 in 4 . 

The tensile unit-stress on the bottom fiber at the section AB, then, is 

Me 198,000X6 
S_ T~ 864 


= 1370 lb. per sq. in. 


provided that this unit-stress is not greater than the proportional limit of the 
material. A compressive unit-stress of the same magnitude occurs on the 
top fiber. 

Second Method of Solution .—Instead of expressing the resisting moment 
as the algebraic sum of the moments of the stresses on the fibers of the cross- 
section, which leads to the expression sl/c, the resisting moment may be 
expressed as the moment of the couple formed by the resultants of the tensile 
and compressive stresses. Thus as shown in Fig. 80 the action line of the 
resultant, C, of the compressive stress is f of OA from O, and similarly the 
resultant T acts at a distance of f OB from O. Hence the resisting moment is 
CX\d or TXfd. Further, the magnitude of C (and T ) is the product of the 
average unit-stress, |s, on the area above (or below) the neutral axis and that 

area. Thus 



Fig. 80.—Resisting couple. 


and hence, 


C = T=hsXhbd, 
and this may be written 

C = T=-X-bd\ 
c 8 

Therefore the resisting moment 

si 2 

M r =-X-bd*X-d 
c 8 3 

bd 3 

12 ’ 

bd 3 si 
12 = ~c’ 


which is the same equation as used in the first method of solution. It is im¬ 
portant to note that the resultants T and C do not act at a distance of f of c 
from O unless the section of the beam is of constant width. 

Problem 77 . —A T-beam (Fig. 81) is subjected to a concentrated load of 
4000 lb. at the center of the span. The beam is made of material having a tensile 


























THE FLEXURE FORMULA 


89 


proportional limit of 4000 lb. per sq. in. and a compressive proportional limit 
of 8000 lb. per sq. in. Find the maximum tensile and compressive unit-stresses 
at the mid-span section, and find the ratios of these stresses to the proportional 
limits. 



Solution .—The distance, y, of the centroidal axis (and hence of the neutral 
axis) from the bottom line of the section is found to be 3 in. Thus (see Art. 
160), 

ay = 2a 0 ?/o 


_ 12X5+12X1 0 

y =-=3 

y 12+12 


in. 


The moment of inertia of the cross-section with respect to the neutral axis 
is found as follows (see Art. 166): 

J x = ts2(6) 3 +12X(2) 2 +tV6(2) 3 +12X(2) 2 

= 136 in. 4 


The bending moment about the mid-section is 


M = R x X- = 2000X72 = 144,000 lb.-in. 
z 


Therefore, the maximum tensile unit-stress (on the bottom fiber of the beam) 
at the mid-section is 

Me 144,000X3 


Sf 


= 3180 lb./sq. in. 


I 136 

and the maximum compressive unit-stress (on the top fiber) is 


5 5 

Sc =- S< = - 3180 = 5300 lb. per sq. in. 
3 3 


The ratio of each stress to the corresponding proportional limit is given 
below. Thus the tensile stress is about 0.8 of the tensile proportional limit 


















90 


TRANSVERSE LOADS 


whereas the compressive stress, although larger than the tensile stress, is only 
0.66 of the compressive proportional limit of the material: 

= 0.795 (tension), = 0.662 (compression) 

This problem shows the advantage of a T-section (or a similar section) for 
material (such as cast iron) that is not equally strong in tension and compres¬ 
sion. 


PROBLEMS 

78. Find the fiber unit-stress at a point on the section AB (Fig. 79) at a 
distance of 2 in. from the top face of the beam. Ans. s = 913 lb. per sq. in. 

79. Find the unit-stress on the bottom fiber of the beam in Fig. 79 on a 
section beneath the concentrated load. 

80 Find the maximum compressive unit-stress on a section of the beam in 
Fig. 81 at a distance of 4 ft. from the left end. 

81. A steel shaft 4 in. in diameter is used as a cantilever beam and loaded 
as shown in Fig. 82. Find the maximum unit-stress in the section, AB, at 
the wall. Ans. s = 6880 lb. per sq. in. 


500 lb. 



1000 lb. 


300 lb. per ft. 




niiiiiiiiiiiii'iTTrm T 

j 2- 

— ft 


w 




~r 

w 

, i 
ZZZZA^,, 


<-4 


Fig. 83. 


82. Find the maximum tensile fiber unit-stress on the section above the 
left support of the cast-iron beam shown in Fig. 83. Find also the maximum 
tensile fiber unit-stress on a section midway between the supports. State 
in each case whether the fiber on which the unit-stress occurs is on the top or 
bottom of the beam. 

35. Section of Maximum Bending Moment.—If a beam has a 

constant cross-section throughout its length, the values of - 

c 

(section modulus) for all sections are equal, and hence the maxi- 

sl 

mum value of s in the flexure formula, M = —, occurs in the sec- 

c 

tion at which the bending moment, M, is a maximum. It is 
important, therefore, to obtain a method of locating the section at 


















SECTION OF MAXIMUM BENDING MOMENT 


91 


which the bending moment is a maximum. This section is called 
the dangerous section of the beam, and may be located in accordance 
with the following statement: 

The section of a beam at which the bending moment is 
maximum is the section at which the vertical shear is either 
equal to zero or changes sign. 

Proof .—In Fig. 84 is represented a beam subjected to a con¬ 
centrated and a uniformly distributed load. The bending moment, 
M X) at the distance x from the left, support is 

M x =R lX -P(x~a)-~. 


Now the value of x that will make M x a maximum is the value 


that will make the first deriv¬ 
ative of M x with respect to x 
equal to zero. The first deriv¬ 
ative of M x with respect to x 
is 

— =Ri-P-wx. (40) 


P 



W lb. per ft. 

iiuuJ 

win 

iiiiiiiiiiiiiiiiiiiiii 

\ x ~^i 

IR. 

tR 


Fig. 84. 


Therefore, the value of x that will make M x a maximum may be 
found from the equation 


Ri — P—wx = 0. 


But Ri — P—wx is the vertical shear for the section at the 
distance x from the left support. Therefore, the section at which 
the moment is maximum is the section for which the vertical shear 
is zero. 

A convenient way of locating the section for which the vertical 
shear is zero (dangerous section) is to draw a shear diagram as 
discussed in the following article. 

Further, the relation between the bending moment and the 


vertical shear found above, namely V = is of great importance; 

it will be discussed in greater detail in connection with shear and 
moment diagrams in the next article. 







92 


TRANSVERSE LOADS 


An alternative method of deriving this equation is as follows: 
In Fig. 85 is shown a part of a beam included between two sections 
a distance dx apart on which a distrib¬ 
uted load acts; the load per unit of 
length being w. All the forces acting 
on this part of the beam are shown in 
Fig. 85, and since these forces hold the 
part in equilibrium, the algebraic sum 
of their moments about any point in 
the plane must equal zero. Hence 
if the point 0 is taken as the mo- 

M+Vdx—wdxXhdx— (M+dM) = 0, 

and since products of differentials may be neglected this equation 
reduces to 

Vdx=dM or V= ^.(41) 

36. Shear and Moment Diagrams.— Shear Diagram. —A shear 
diagram for a beam is a curve in which the abscissas represent 
distances along the beam and the ordinates represent the vertical 
shears for the sections at which the ordinates are drawn. 

For example let it be required to draw a shear diagram for the 
beam shown in Fig. 86(a). The reactions are found to be 

Ri = 7000 lb. and R 2 = 5000 lb. 

Now the vertical shear for section A (just to the right of the left 
support) is Va = Ri = 7000 lb. upward; it is represented by AE 
in Fig. 86(6), an upward shear being considered positive and 
plotted above the base line. 

The vertical shear for section C (just to the left of the load 
P) is 

V c = 7000 - 4 X 500 = + 5000 lb. 

and the vertical shear for any section between A and C at the 
distance x, say, from A is 


wdx 






0 



*dx> 


V 

'V+dV 


M + dM 


Fig. 85. 

ment center we have, 


F* = 7000—5002 . . 


(42) 















SHEAR AND MOMENT DIAGRAMS 


93 


and hence the vertical shear decreases with x at the constant rate 
of 500 lb. per foot of length of the beam from 7000 lb. at A (x = 0) 
to 5000 lb. at C (x = 4) as shown in Fig. 86(6). 


P= 4000 lb. 



Fig. 86.—Shear and moment diagrams. 


The vertical shear at section D (just to the right of the con¬ 
centrated load) is 

Fi> = 7000-500X4-4000 = + 1000 lb. 

Thus, there is an abrupt change in the vertical shear in passing 
from sections C to D due to the concentrated load, but the shear 






























94 


TRANSVERSE LOADS 


does not (in this particular problem) pass through zero under the 
load; the shear must, therefore, be equal to zero for some section 
to the right of the concentrated load. Now, the vertical shear for 
any section between D and B at a distance x from the left sup¬ 
port is 

V x = 7000—4000—500x,.(43) 

and hence the vertical shear decreases at the constant rate of 500 
lb. per foot of length of the beam from 1000 lb. at D to —5000 lb. 
for a section just to the left of the right reaction. At 6 feet from 
the left support the vertical shear is zero and hence the bending 
moment at this section is maximum. 

It is important to note that if a concentrated load acts on a 
beam in addition to a distributed load the shear may pass through 
zero under the load, and if it does the location of the dangerous 
Section cannot be found by equating any expression such as 
equations (42) and (43) to zero and solving for x. It is desirable, 
therefore, in finding the dangerous section, to plot a shear diagram. 
If a beam is acted on by concentrated loads only, the dangerous 
sections will always occur under one of the loads since the shear 
changes only at the sections where the loads act. 

Moment Diagram .—A moment diagram for a beam is a curve in 
which the abscissas represent distances along the beam and the 
ordinates represent the bending moments at the sections at which 
the ordinates are drawn. 

Let it be required to draw a moment diagram for the beam 
shown in Fig. 86(a). The bending moment at the section A is 
equal to zero since the moment arm of Ri about A is zero. The 
bending moment at any section between A and (7, at a distance x 
from the left support, is 

M x =R lX -^f } .(44) 

in which x can not have a value greater than 4. The bending 
moment at any section between D and B (x greater than 4), at a 
distance x from the left support, is 



SHEAR AND MOMENT DIAGRAMS 


95 


in which x cannot have a value less than 4. The value 
of the bending moment at the section beneath the load 
(x=A) may be found from either equation by making x equal 
to 4. 

If various values of x from 0 to 4 be substituted in equation 
(44) and the resulting values of the bending moments plotted, the 
moment diagram obtained is AF (Fig. 86c), and if values of x 
from 4 to 16 be substituted in equation (45) the resulting moment 
diagram is FB (Fig. 86c). 

It will be noted that the maximum ordinate to the moment 
curve occurs at the section for which the shear is zero. Further, 
it is important to observe that the moment diagram is composed 
of two distinct curves, AF and FB, which have only one point, F, 
in common, and that the maximum bending moment can be found 
from the equation of only one of these curves. Thus in the beam 
of Fig. 86(a) the value of x for which the bending moment is maxi¬ 
mum can not be substituted in equation (44), since 4 is the greatest 
value x can have in this equation. If values of x greater than 4 
are substituted in equation (44) the corresponding values of M x 
will have no physical meaning; and if values of x less than 4 are 
substituted in equation (45) the corresponding values of M x will 
have no physical meaning. 

Sign of Bending Moment .—The bending moment at any section 
of a horizontal beam is considered to be positive when it produces 
compressive stress on the top fibers of the beam at the section and 
tensile stress on the bottom fibers. Thus, if the bending moment 
is obtained from the forces that lie to the left of the section the 
bending moment is positive when it is clockwise, and if obtained 
from the forces to the right of the section the bending moment is 
positive when it is counterclockwise. 


ILLUSTRATIVE PROBLEMS 

Problem 83.—A solid cylindrical steel shaft is used as a simple beam over 
a span of 12 ft. and is loaded as shown in Fig. 87(a). Draw the shear and 
moment diagrams and find the diameter of the shaft if the working fiber 
unit-stress is 16,000 lb. per sq. in. 

Solution .—The reactions are found to be 


= 6883 lb. and R 2 = 5717 lb. 


98 


TRANSVERSE LOADS 


The shear and moment diagrams are shown in Fig.^87(6). The dangerous 
section is under the concentrated load and the maximum moment is 



M = 6883X7—800X7X1 
= 28,580 lb.-ft. 

The maximum fiber unit-stress is 


28,580 X12 X- 
Mc 2 


S = T = 


oT 


Hence, 


„ 28,580X12X32 . 

d 3 = -———-=218 in. 3 

16.000 X*- 

whence 

d = 6.02 in. 


Problem 84 . —Show that the maximum bending moment for a simple 
beam loaded as shown in Fig. 88(a) is 

M = | Wl 

in which W is the total load on the beam and l is the span of the beam. 

Solution .—From symmetry it is known that the shear is zero at the center 
of the span and that each reaction equals W /2. Let the intensity of the load 
at the center be w c lb. per ft. Now the resultant, R, of the load to the left of 
the mid-span section is equal to the average load per foot (w c / 2) times the 
length (/). But the action line of R is at the distance of f Xl/2 from the left 
support. Thus the distributed load to the 
left of the center of the beam may be re¬ 
placed by R as shown in Fig. 88(6) in get¬ 
ting the moment about the mid-span sec¬ 
tion. The bending moment about the 
mid-span section then, is 

J- W L-^l x iL 

• 2 2 2 2 32 

But, the average load per foot 


the span length ( l ) equals W, 
Hence, 


^ ) times 





l Wl 1 

M = W - = -Wl. 

4 12 6 
























































SHEAR AND MOMENT DIAGRAMS 


97 


PROBLEMS FOR ARTICLES 34 TO 36 

85 to 90. Verify the shear and moment diagrams (shown in Figs. 89 to 94); 
also verify the expressions for the vertical shear and the bending moment at a 
section at the distance x from the left end of the bean, and the expressions]for 
the maximum moment, as given in Figs. 89 to 94. 


Prob. 85. 


W 



M x =-Wx, M max =-Wl. 
Fig. 89. 


Prob. 86. 



Fig. 90. 


Prob. 87. 



. W w W , 
Vx=-^-W= when 

tO /. 

w 

F m ax. = ±^-. 

w i 

M x = yx, when x = 0 to 


W 


M x = — W U-o) . when 


x= \ t0 1 


M — — — x Wl 

M max. — ~2 2 — 4 W 

Fig. 91. 



V x ="2 ~ wx . W = wl. 
_ W 

' max. -L n • 


W wx 2 
M x ^x 2 ’ 

Wl Wl 1W7 , 72 

Mmax . =“ 2*2—2 4 = ^ = ^ 

Fig. 92. 












































98 


TRANSVERSE LOADS 


Prob. 89. 
W 



when x = 0 to a. 

7 ™and 

— W(x—a) 
when x = a to l. 

T . _ Wa _Wb 

V max. — > ^max. ——j—a. 

Fig. 93. 


Prob. 90. 



W 

Vx = 0, and M*=—a, 

when x = a to (a+6). 

W 
2 

TV 

and M* = IFa —jr (x — b) 




when 


z = (a+5) to i. 
W 
= 2 

Fig. 94. 


91. Plot a shear diagram and find the dangerous section for the beam shown 
in Fig. 95. Find also the maximum bending moment and the maximum fiber 
unit-stress in the beam. 

92. Draw to scale a shear and a moment diagram for the beam shown in 

Fig. 96. If the beam is a cylindrical bar what diameter should it have if the 
working fiber stress is 16,000 lb. per sq. in.? Ans. d = 4.97 in. 


400 lb. per ft. 

2000 lb. 

<-3-—>| 

min 1111111111 nr 

lllllll 

l 


J* 


' Fig. 95. 


4000 

L' Q N Ilf d / V 

Hb. 

s A N. 

rx o .>rx > 

500 lb. per ft. 1 

s- *± -^ 

rmriiiimmiiiii 

1 

16 >| 



Fig. 96. 


93. A simple beam having a span of 12 ft. is made of yellow pine for which 
the working fiber stress is 1000 lb. per sq. in. The beam carries concen¬ 
trated loads of 2000, 4000 and 6000 lb. at distances of 3, 6 and 7 ft., respectively, 
from the left support and also a total uniformly distributed load of 1200 lb. 
If the width of the beam is 10 in. what should be the depth ? 

Ans. d = 15.2in. 

94. A freight-car axle (Fig. 97) is 4 in. in diameter. The distance from the 
center of an axle box to the action line of the rail pressure is 8 in. and the 














































PROBLEMS 


99 


distance between the rail pressures is 4.90 ft. Find the maximum fiber 
unit-stress in the axle if each rail pressure is 5 tons. 




95. The steel crankpin (Fig. 98) of a steam engine has a force of 25,000 lb. 
transmitted to it by the connecting rod. The dimensions of the pin as shown 
in Fig. 32 were determined from the allowable bearing pressure. Find the 
allowable bearing pressure, and the fiber unit-stress. The load may be 
assumed to be uniformly distributed over the length of the pin. 

Arts. s& = 800 lb. per sq. in., s = 6360 lb. per sq. in. 

96. A simple beam having a span of 16 ft. carries a total uniformly dis¬ 
tributed load of W lb., and a concentrated load, equal to IF/2, at the center of 
the span. If the cross-section of the beam is rectangular, 6 in. wide by 12 in. 
deep, and the allowable fiber unit-stress is 800 lb. per sq. in., find the value 
of W. 

97. The cast-iron frame shown in Fig. 99 is subjected to a load P of 4000 

lb. Find the maximum tensile and compressive fiber stresses at the section 
AB. Ans. i/i = 3.5 in., 7 = 392 in. 4 , st = 857 lb. per sq. in., 

s c = 1590 lb. per sq. in. 




Fig. 100. 


98. What force P (Fig. 100) will cause a maximum fiber unit-stress of 
16,000 lb. per sq. in. in the I-beam ? 

















































100 


TRANSVERSE LOADS 


99. The load P (Fig. 101) is 1000 lb. What should be the depth d of the 
timber beam A if the working unit-stress in the material is 1200 lb. per sq. in ? 

Ans. <2 = 9.5 in. 




0-S' • 

A v 

d 


£ ,B rf_f 


Fig. 102. 


100. The simple timber beam shown in Fig. 102 has a hole 4 in. in diameter 
bored through the beam, the center line of the hole lying in the neutral surface 
and in the section AB, 3 ft. from the left support. The beam is subjected to 
a load P of 5000 lb. Find the maximum unit-stress that occurs in the section 
AB and also in the section for which the vertical shear is zero. The beam has 
a rectangular section 4 in. wide and 12 in. deep. 

101. Sand is piled on a floor so that the load on the beams supporting 

the floor varies from zero pressure at one end 
to a pressure of w lb. per ft. at the other end 
(Fig. 103). Show that the vertical shear is 
zero for a section at a distance of 0.59 1 from 
the left support, and that the maximum bend¬ 
ing moment is 

Fig. 103.—Non-uniformly di.s- iq/ 

tributed load. ‘ M =-^=Wl. =0.128 Wl . . (47). 

102. Draw the shear and moment diagrams for the beam shown in Fig. 104. 

Find the maximum unit-stress in the beam if the beam has a rectangular 
cross-section 4 in. wide by 8 in. deep. Ans. $ = 983 lb. per sq. in. 



2000 lb. 


2000 lb. 



1000 lb. 



Fig. 104. 


Note.— The following problems are to be solved with the aid of a steel 
company’s handbook. Find the missing terms in the table. All the 
beams are simple beams and all distances are measured from the left 
support. 


v «*> 







































RELATION BETWEEN SHEAR AND MOMENT 


101 


Prob. 

No. 

Length 
of Span 
(feet) 

Uniform 

Load 

(lb./ft.) 

Concentrated 

Load 

(lb.) 

Distance to 
Concentrated 
Load (ft.) 

I-Beam Size 

103 

10 

2000 

0 

0 

? 

104 

12 

? 

0 

0 

10" 25-lb. 

105 

14 

0 

28000 

7 

? 

106 

14 

0 

? 

7 

9" 21-lb. 

107 

16 

1000 

10000 

8 

? 


37. Relation between Shear and Moment. —In Art. 35 it was 
shown that 



which states that the vertical shear for a section of a beam is equal 
to the rate, with respect to the distance along the beam, at which 
the bending moment is changing at the section. Or, the equation 
may be written 

dM = Vdx, 


which states that the difference, dM (Fig. 105), between the bend¬ 
ing moments at two sections that are the distance dx apart is equal 
to the area, Vdx, under the shear 


curve between the two sections. 
Now the difference (M 2 —Mi or 
AM) between the bending moments 
at the sections X 2 and x\ would be 


A M = 


ru 2 

JMy 


dM 


(Fig. 105). But, from the above 
equation, 



Fig. 105.—Relation between bend- 
(48) ing moment and vertical shear. 


r 

dM = 

JMi Jx 1 

rx 2 

And I Vdx represents the area under the shear curve between 

J x 1 


the ordinates x\ and X 2 , and hence. 


M 2 —Mi = area under the shear curve between ordinates 
xi and X 2 . Therefore, 





















102 


TRANSVERSE LOADS 


the difference between the bending moments at two sections 
of a beam is represented by the area under the shear curve 
between ordinates at the two sections. 



Fig. 106.—Shear diagram. 


Thus, the bending moment at a given section of a simple beam 
is represented by the area under the shear curve between the end 

of the beam (M = 0) and the given 
section. For example, let it be 
required to find the maximum 
moment for a simple beam sub¬ 
jected to a total load W, uniformly 
distributed over the entire span, 1. 

The shear diagram is shown in 
Fig. 106. According to the above proposition the difference 
between the bending moments at the center and end is represented 
by the area under the shear curve between these two sections. 
But the moment at the end is zero and hence the moment at the 
center is 

M c = \ base X altitude 
2 2 X 2 8 ’ 


which checks the result found in Prob. 88. 


PROBLEMS 

108. Show, by the above method, that the maximum bending moment for 
a simple beam carrying a concentrated load P at mid-span is, M = \Pl. 

109. Show, by the above method, that the bending moment at any section 
within the middle third of a beam when loaded with equal concentrated loads 
at the third-points is, M = %Pl. 

110. Find, by the above method, the maximum moment for the beam 
shown in Fig. 93. 

38. Overhanging Beams. —The discussion in the preceding 
articles concerning the stresses developed in simple and cantilever 
beams also applies to overhanging beams. The bending moment 
at a section in an overhanging beam, however, may be negative, and 
there are two maximum moments to be considered—the maximum 
positive moment and the maximum negative moment. In other 
words, the vertical shear passes through zero at two or more 
sections of the beam. Further, since the moment changes from a 






OVERHANGING BEAMS 


103 


positive to a negative value it must be zero at some section; 
the section at which the bending moment is zero is called the 
point of inflection. 

The above facts will be discussed in connection with the follow¬ 
ing illustrative problem. 

ILLUSTRATIVE PROBLEM 

Problem 111. —A 6-in. by 12-in. by 14-ft. timber beam is supported and 
loaded as shown in Fig. 107(a). (a) Draw the shear and moment diagrams. 

( b ) Find the position of the dangerous section. (c) Find the maximum 
positive moment. ( d ) Find the maximum negative moment, (e) Find the 
point of inflection.' (/) Find the maximum fiber unit-stress. 

Solution. —The reactions are found to be 

#i = 4600 lb. and #> = 7400 lb. 

(a) The shear and moment diagrams are shown in Fig. 107(6). The 
vertical shear is equal to zero for a section 3.25 ft. from the left support, and 
passes through zero at the section 
above the right support. These 
two sections, then, are the danger¬ 
ous sections. The bending mo¬ 
ment at any section to the left of 
the first concentrated load (Pi), 
at a distance x from the left sup¬ 
port, is 

M x = 4600X-800-, . (49) 

in which x has any value from 0 
to 3. The bending moment at 
any section between Pi and the 
right reaction is 

x 2 

M x = 4600.r—800— 

A 

—2000(x—3), (50) 

in which x cannot have a value 
less than 3 ft. or greater than 10 
ft. The bending moment at any section between the right reaction and the 
right end of the beam is 

ikf x = 4600x+7400(x-10)-2000(x-3)-8000(a;-5), . . . (51) 

in which x cannot be less than 10 or greater than 14 ft. Or, if x is measured 
from the right end of the beam instead of the left end then the bending moment 
at any section between P 2 and P 2 is 

M x = —2000a:. 


P, = 2000 lb P 2 = 2000 lb. 



(52) 


































104 


TRANSVERSE LOADS 


Equations (51) and (52) will give, of course, the same value for the bending 
moment at a given section of the beam. 

(6 ) The maximum positive moment is 

800X(3.25) 2 

Mpos. = 4600 X3.25 - 2000 X 0.25--- 

A 

= 10,230 lb.-ft. 

(c) The maximum negative moment, obtained from the forces to the left 
of the section, is 

M neg = 4600 X10 - 2000 X7 - 800 X10 X 5 
= -8000 lb.-ft. 

Or, when obtained from the forces to the right of the section, it is 
M neg = -2000X4= -8000 lb.-ft. • 

( d ) The bending moment changes signs at a section between the dangerous 
sections. Hence, to locate the inflection point equation (50) may be equated to 
zero. Thus, 

3/2 

M x = 4600a: -800- - 2000(x-3)= 0. 

Therefore, 

x = 0 or 8.63 ft. 

and hence the inflection point is 8.63 ft. from the left support. 

(e) Since the maximum positive moment is larger than the maximum 
negative moment, the maximum fiber unit-stress is 

Me 10230X12X6 
S ~~T~ T26X(12) 3 

= 852 lb. per sq. in. 


PROBLEMS 


For each of the beams described in the following problems obtain: (a) The 
shear and moment diagrams; (6) the maximum positive and negative bending 
moments; (c) the points of inflection; (d) the maximum fiber unit-stress in 
the beam. The weight of the beam may be neglected in each problem. 


112. The beani shown in Fig. 108 has the following dimensions: 6 = 8 in., 
d = 10 in., and the loads are, P = 1000 lb. and w = 300 lb. per ft. 

Ans. (d) s = 1050 lb. per sq. in. 



P 

W lb. per ft. 

mn 

n mm i m iTimT 

mnraT 


H- 

--—16-- 

—4—> 


□$ 

T 



Fig. 108. 


Fig. 109. 


113. The diameter, d, of the beam shown in Fig. 109, is 4 in. and the loads 
are P x = 1500, P 2 = 6000 lb., and P 3 =3000 lb. 


CO?* 



















ECONOMICAL SECTIONS OF BEAMS 


105 


114. See the beam shown in Fig. 73, and consider the beam to have a 
rectangular cross-section 4 in. wide and 12 in. deep. 

39. Economical Sections of Beams. —The efficient use of ma¬ 
terial in a force-resisting member requires (a) the selection of 
available material that is well suited to the use of the member, and 
(6) the distribution of the material in the member so that it can 
best resist the forces acting on the member. The use of material, 
however, must always be considered in relation to the cost involved. 

The selection of engineering materials best suited to various 
purposes is discussed in Part II. We are here interested mainly 
in the effect of the distribution of the material in a beam on the 
resistance of the beam to external loads; this involves (a) the 
effect of the shape of cross-section and (6) the effect of a variation 
of size of cross-section along the beam. 

(a) Effect of Shape of Cross-section .—If a beam having a con¬ 
stant cross-section safely resists static loads the maximum fiber 

T 

L 

Fig. 110.—Sections having large values of 

unit-stress at the dangerous section must not exceed the allowable 
or working unit-stress that experiments and experience have shown 

Me 

to be permissible. That is, s in the flexure formula, s = ~ff> must 

not be greater than the working unit-stress when M is the maximum 
bending moment. 

The flexure formula shows that, when a given unit-stress s is 
developed in the beam, the bending moment M required to develop 

this unit-stress is large when - is large. Now - is made large by 

c c 

forming the cross-section so that the greater part of the area is as 
far from the neutral axis as practicable. Thus, steel beams are 
rolled in the form of I-sections (Fig. 110a), channel-sections (Fig. 
1106), etc.; further, built-up steel beams of various shapes (Fig. 
110c and d), and also cast-iron beams are made to conform to this 
principle. Steel beams are made with the two flanges equal in 
area since the proportional limits in compression and tension are 
















106 


TRANSVERSE LOADS 


approximately equal. Cast-iron beams, however, are cast with 
the tensile flanges larger in area than the compressive flange (see 
Fig. 99) since the compressive strength of cast iron is much larger 
(about four times) than the tensile strength. Although cast-iron 
beams are rarely used in buildings or bridges, they are frequently 
used in machine frames that are subjected to bending. 

(b) Effect of a Varying Section Along the Beam .—As discussed 
in Art. 36 the bending moment, in general, varies along a beam and 
is maximum at one section of the beam. If, then, a beam has a 


constant cross-section 


(ff c = Si constant 


the maximum fiber unit- 


stress will occur on the outer fiber of the section at which the 



Fig. 111.—Beams that approach the conditions for uniform stress. 


bending moment is maximum, and the unit-stress in the outer 
fibers at all other sections will be less than that at the dangerous 
section. Therefore, when the beam is carrying the load that causes 
the allowable fiber unit-stress in the beam there is much material 
in the beam on either side of the dangerous section that is under¬ 
stressed, and hence this under-stressed material could be saved by 

varying the cross-section of the beam so that the ^ would vary as 
the bending moment M varies. This would cause the unit-stresses 
in the outer fibers of all sections to be equal, since s = ~. 

c 

Rolled steel beams nearly always are beams of constant cross- 
section since the cost of rolling a beam of variable cross-section 




























SHEARING STRESS IN A BEAM 


107 


would offset the saving in material. Built-up beams, such as leaf 
springs (Fig. 111a) and plate girders (Fig. 11 Id), frames of cars 
(Fig. llle), forged axles (Fig. 1116) and turned axles (Fig. 111c), 
etc., are frequently made with a variable section so that the beam 
approximates a beam of uniform strength. 

40. Shearing Stress in a Beam— In Art. 33 it was shown that 
in a section of a beam there is shearing stress as well as tensile and 
compressive stresses, and that the total resisting shearing stress, 
V r , in any section is equal to the vertical shear, V, for the section. 
That is, V=V f . In the preceding articles the tensile (or compres¬ 
sive) unit-stress at any point in the beam was found in terms of 
the external loads and the dimensions of the beam, and the problem 
now to be considered is that of expressing the shearing unit-stress 
at any point in a beam in terms of the external loads and the 
dimensions of the beam. 

If the shearing unit-stress on the cross-sectional area, a, of a 
beam were constant and equal to s s the resisting shear, V r , would 
be equal to as s , and hence the shearing unit-stress at any point of 



the area would be equal to — l s s = — j. But, the shearing unit- 


stress is not constant over the area, as will be shown later in this 

V 

article, and hence the equation s s =— gives only the averages hearing 

unit-stress. Shearing stress in beams is of importance mainly in 
timber beams, concrete beams, and some built-up steel beams. 

As stated in Art. 17, if a shearing unit-stress occurs on one 
plane at a point in a body, a shearing unit-stress of equal magni¬ 
tude must occur at the same point on another plane at right angles 
to the first plane. Thus, in Fig. 112, X represents a small block 
in a beam; there is a vertical upward shearing stress on its left face 
and a vertical downward shearing stress on its right face, but these 
two forces form a couple which would rotate the block and hence, 
since the block is in equilibrium, there must be horizontal shearing 
forces on its upper and lower faces as indicated in Fig. 112. 
Thus we are led to the conclusion (see Art. 17 for proof) that the 
vertical and horizontal shearing unit-stresses at any point are 
equal. 

The horizontal shearing unit-stress (and also the vertical 
shearing unit-stress) at any point in a beam may be found, in terms 
of the external forces and the dimensions of the beam as follows: 


108 


TRANSVERSE LOADS 


Let a block, B, having a small width dx (Fig. 112) be removed 
from the beam and be replaced by the forces that the block was 
exerting on the beam. These forces must have been exerted, of 
course, on the faces that were in contact with the beam. Now, as 
shown in Art. 34, the compressive (or tensile) unit-stress at any 
point on any cross-section of the beam varies directly as the 
distance of the point from the neutral axis, being zero at the 
neutral axis. Thus the block B must have pushed horizontally at 
its two end faces as shown in Fig. 112(a) (or the beam must-have 
pushed on the end faces of the block as shown in Fig. 1126). 

But since the bending moments at the two sections are not, 
in general, equal, the sum of the stresses on the two faces are not 
equal and hence there must be a shearing stress on the bottom face 
of the block. 


( 6 ) 




Thus, if H' and H represent the sum of the stresses on the 
right and left faces, respectively, the total shearing stress on the 
bottom force is equal to H'— H, assuming H' to be larger than 
H. Further, since the area dx-t on which the shearing stress 
occurs is small the shearing unit-stress, s s , may be assumed to be 
constant over the area and hence the total shearing stress is s s • dx ■ t. 
Therefore, 

s s dx-t = H' — H. 

Now H r = I Sy'da in which s r y is the unit-stress in the plane FG 
Jv 0 

on a fiber at the distance y from the neutral axis (Fig. 112a). 
s ^ s^ 

Further, — = -=^=a constant (Art. 34.) Therefore, 
y c 



yda and similarly 




















SHEARING STRESS IN A BEAM 


109 



moments at the right and left sections respectively. Therefore, 



But M'—M = dM since the two sections are the small distance 
dx apart. Hence 



and from Art. 37, 



dx 


Therefore, 



(53) 


in which s s is the horizontal shearing unit-stress (and also the 
vertical shearing unit-stress) in a cross-section for which the 
vetical shear is V, and at a point whose distance from the neutral 


axis is y 0 , the thickness of the beam at the distance y 0 from the 


neutral axis being t; 7 is the moment of inertia of the whole cross- 
section of the beam about the neutral axis. The expression 



yda is the first moment (often called the statical moment), 


about the neutral axis, of that part of the cross-sectional area of 
the beam between the plane on which the horizontal shearing 
unit-stress occurs and the outer face of the beam (that is, between 
y 0 and c ). This area is the cross-hatched area a' in the end view 
in Fig. 112a. Further, if the distance, y, of the centroid of the 
area a' from the neutral axis is known, the moment of this area 
may be found from the product a'y since 


Hence the above equation may be written 



yda = a'y (Art. 158). j 





110 


TRANSVERSE LOADS 


The above equations are valid only when the tensile or com¬ 
pressive unit-stress in the beam does not exceed the proportional 
limit of the material since the flexure formula, which is based on 
this assumption, was used in its derivation. 


It is important, next, to locate the point in a beam at which the 
shearing unit-stress, s s , is a maximum. If the beam has a constant 
cross-section, I and t are constant, and hence the maximum value 
of s s in equations (53) and (54) will occur in the section of the beam 
for which V is maximum; for a simple beam subjected to a uniform 
load V is maximum close to one support. Further, in any section 


of constant thickness, 


will te maximum when I yda or a'y is 



Fig. 113.—Variation’of shearing stress in length of beam. Fig. 114.—Varia¬ 

tion of shearing 
stress over sec¬ 
tion of beam. 

maximum, which occurs when y 0 is zero, that is, s s is maximum 
at the neutral surface. Thus in a simple beam of constant thickness , 
subjected to a uniform load the horizontal shearing unit-stress 
varies throughout the beam as shown in Fig. 113, and at any point 
in a section the horizontal and vertical shearing unit-stresses are 
equal as indicated in Fig. 114. 

Thus, the maximum shearing unit-stress in any section of a 
beam having a rectangular cross-section hd (Fig. 113) is 

V - V ,dd 

s ‘ = Tt a ' y = ^bcp-- V 6 2 X 4 

= 3 V = 3F 
2 hd 2 a’ 

in which a is the area of the whole cross-section. Hence the 
maximum value s a in any section of a rectangular beam is 50 











































SHEARING STRESS IN A BEAM 


111 


per cent greater than the average shearing unit-stress in the sec¬ 
tion. If a beam has a circular cross-section the maximum value 

of is 33J per cent greater than —, that is, 

CL 

4 V 

Ss = o —• 

3 a 

ILLUSTRATIVE PROBLEMS 

Problem 115. —Find the maximum shearing unit-stress in a 15-in. 45-lb. 
channel (Fig. 115a) when used as a simple beam on a 12-ft. span and subjected 
to a concentrated load of 20,000 lb. at the center of the span. (Neglect the 
weight of the beam.) 

Solution .—A steel maker’s handbook gives the following values in addition 
to those shown in Fig. 115(a). 



I a; = 375 in. 4 a = 13.2 sq. in. 

The maximum shearing unit-stress is 
1 " 

s,=yay 

V = 10,000 lb. /=375 in 4 . ( = 0.62 in. 

The value of a'y is most easily found as the sum of the moments of the areas 
ai and a 2 (Fig. 1156). Thus, 

a'y=axyi+a 2 y2 

= (3X0.65)X^7.5-^j+(0.62X7.5)X3.75 
= 31.5 in 3 . 


10,000 31.5 
375 *0.62 

= 1350 lb. per sq. in., 


Therefore 














112 


TRANSVERSE LOADS 


and hence the maximum shearing unit-stress in the beam is 1350 lb. per sq. in., 
provided that the tensile or compressive unit-stress does not exceed the elastic 
limit of the material. 

The maximum tensile or compressive unit-stress is 


Me 10,000X72X7.5 
I ~ 375 


= 14,400 lb. per sq. in. 


which is less than the elastic limit of the material. 

The average shearing unit-stress is 

V 10,000 

Sa ' , = a = l3^i = 755 1b - Per8q - *“• 


and hence the maximum shearing unit-stress is 1.79 times the average. 


Shearing Stress in Web of I-beam .—It is customary in struc¬ 
tural design to assume that the flanges of a channel-beam or an 
I-beam are effective in resisting bending but not in resisting shear, 
and the maximum shearing unit-stress in the beam is assumed to 
be equal to the average shearing unit-stress over the web only, 
regarded as extending the entire depth of the beam. Thus in the 
above problem 

Wav. m web = td = 0 62 X15 = 1074 lb ' per sq * m ' 


This method, therefore, gives only approximate results for the 
channel section, although in the case of an I-section the method 
yields results somewhat closer to those found by the correct 
method. 

The justification for assuming the shearing unit-stress in an 
I-beam to be distributed uniformly over the web considered 

to extend the entire depth of the 
beam may be found in a study 
of Fig. 116. In any section of an 
I-beam s s , according to equation 
(54), will be large where i is small, 
and small where t is large, the 



■sAa 


Fig. 116 . —Distribution of shearing 
stress over section of I-beam. 


effect on s 8 of the term yda be¬ 


f 


ing relatively small in the case 
of an I-section; further, s s will change abruptly where t changes 
abruptly. Hence, the horizontal (and vertical) shearing unit 
stress at a section in an I-beam varies as shown in Fig. 116(6), and 
















SHEARING STRESS IN A BEAM 


113 


the total shearing stress (s a -da) on the elementary strips of the 
section (the total stress per unit of depth) is approximately con¬ 
stant for the whole depth of the section as shown in Fig. 116(c). 
Therefore, the method, stated above, for finding the shearing unit- 
stress in the web of an I-beam yields results that contain small 
errors. 

PROBLEMS 

116. A simple beam made of wood has a cross-section 6 in. wide by 8 in. 
deep. A concentrated load of 16,000 lb. acts at the mid-span section, and the 
span is 10 ft. (a) Find the maximum shearing unit-stress. ( b ) Find also 
the shearing unit-stress at a point 2 in. from the top of the beam in a section 
2 ft. from the left support. 

Ans. (a) s s = 250 lb. per sq. in. (6) s s = 188 lb. per sq. in. 

117. A simple timber beam has a hollow square cross-section; the outside 
dimensions are 8 in. by 8 in. and the inside dimensions are 4 in. by 4 in. The 
span of the beam is 10 ft. and the beam is subjected to a uniformly distributed 
load of 1500 lb. per ft. Find the maximum shearing unit-stress. Where in 
the beam does this stress occur ? 

Ans. s s = 328 lb. per sq. in. 

118. A Douglas fir beam (Fig. 117) having a cross-section 8 in. by 16 in. 

when tested as a simple beam with a span of 15 ft. by equal concentrated loads 
at the third points, failed by shear (as shown in Fig. 117) when the total load 
on the beam was 48,850 lb. Find the maximum shearing unit-stress in the 
beam when failure occurred. Ans. s s =2SS lb. per sq. in. 



Fig. 117.—Timber beam that failed by horizontal shear. 


119. Calculate the maximum shearing unit-stress in a 15-in. 60-lb. I-beam 
having a span of 12 ft. when loaded with a uniformly distributed load which 
causes a maximum tensile unit-stress of 16,000 lb. per sq. in. 

Ans. $ s = 4800 lb. per sq. in. 







114 


TRANSVERSE LOADS 


120. A large Douglas-fir beam is to be designed to resist a concentrated 
load of 20,000 lb. at the center of the span. Use a working stress equal to 
one-third of the ultimate shearing unit-stress found in Problem 118 (288 lb. 
per sq. in.). If the span of the beam is 12it. and the width of the beam is to 
be 8 in. what should be the depth of the beam ? A working shearing stress 
of 100 lb. per sq. in. is frequently specified for large Douglas-fir beams, and 125 
lb. per sq. in. for dense pine. 

41. Stress Beyond Proportional Limit. Modulus of Rupture.— 

si 

As shown in Art. 34 the value of 5 in the flexure formula, M = —, is 

c 

the tensile or compressive unit-stress in the beam only when s does 
not exceed the 'proportional limit of the material. If the bending 
moment M causes a unit-stress greater than the proportional 
limit of the material the bending moment is still held in equilibrium 



Fig. 118.—Change in distribution of stress when proportional limit is 

exceeded. 


by the resisting moment but the resisting moment is not given by 
si 

the expression — since this expression was found by assuming that 

the unit-stress at any point varies directly as the distance of the 
point from the neutral axis; and this is true only when the maxi¬ 
mum unit-stress does not exceed the proportional limit. 

The distribution of stress on the cross-section of a beam when 
the maximum unit-stress on the section is less and greater than the 
proportional limit are shown in Figs. 118(a) and (6), respectively, 
and may be explained as follows: Experiments have shown that a 
plane section of a beam remains approximately plane after the beam 
is stressed even though the unit-stress exceeds the proportional 
limit of the material, and hence the unit-strain at any point is 
proportional to the distance of the point from the neutral axis in 
both beams of Fig. 118. In the beam that is stressed beyond the 
proportional limit (Fig. 1186) the unit-stress is not proportional 
to the unit-strain except near the neutral axis where the strain is 














STRESS BEYOND PROPORTIONAL LIMIT 115 

small, and hence the unit-stress does not vary directly as the dis¬ 
tance from the neutral axis except for short distances from the 
neutral axis; when the unit-stress on any fiber exceeds the propor¬ 
tional limit the unit-stress on that fiber increases at a less rate than 
does the strain; the curve showing the distribution of unit-stress 
on the section is the same as a portion of the stress-stress diagram 
for the material. 

Modulus of Rupture .—From Fig. 118 it is evident that with a 
given unit-stress in the outer fiber of the beam the resisting 
moment is larger when the unit-stress varies as shown in Fig. 118(6) 
than it would be if the unit-stress varied directly as the distance 
from the neutral axis. Therefore, the resisting moment in a beam 
when stressed to a given unit-stress beyond the proportional limit 

si 

is larger than that found from the expression by substituting the 

given unit-stress for s. Thus, if the ultimate strength of the 

si 

material is substituted for s in the flexure formula M = — } the 

c 

resulting value of M will be less than the maximum moment that 
the beam can resist. 

Tests of beams of various material have shown that the maxi¬ 
mum bending moment a beam can resist may be from 20 to 100 
per cent greater, depending on the material and the shape of cross- 
sections, than the value of M found from the flexure formula by 
substituting the ultimate strength of the material for s. 

If, then, the allowable or working unit-stress in a beam is taken 
to be some proportion (i, say) of the tensile or compressive 
ultimate strength (using the lesser value) the maximum bending 
moment that could be applied to the beam would be greater than 
five times the moment that would produce the working stress. For 
some materials (particularly brittle materials) a working unit- 
stress based on the ultimate strength of the material seems un¬ 
necessarily small and this fact is sometimes employed to justify 
larger working stresses in beams than is used for tension members. 
The value of s found from the flexure formula by substituting for 
M the value of the maximum bending moment that a beam re¬ 
sists when tested to rupture is called the modulus of rupture of 
the material in flexure. Thus, if it is denoted by s r , we have 

_ M max.C 

Sr — j 


(55) 




116 


TRANSVERSE LOADS 


It should be noted that s r is not the unit-stress in the material 
caused by the bending moment M m &x ; it is not the ultimate 
strength of the material. 

But although the modulus of rupture is not the actual maxi¬ 
mum unit-stress in the beam the relative values of the actual unit- 
stresses in two beams when stressed beyond the proportional 
limit may be found from the moduli of rupture provided that the 
two beams have cross-sections of the same shape; and only slight 
error will occur if the beams have cross-sections of different shapes 
provided that the cross-sections are symmetrical with respect to 
the neutral axis and that the stress-strain curves of the material 
in tension and compression have the same slopes (moduli of elasti¬ 
city in tension and compression are equal). If these two conditions 
are not satisfied the neutral axis shifts when the beam is stressed 

beyond the elastic limit and 
does not pass through the cen- 
troidal axis of the section. 

For example, in a beam 
with an unsymmetrical section 
(Fig. 119a), the most remote 
fibers on one side of the neu¬ 
tral axis will be stressed to 
their proportional limit be¬ 
fore those on the other side, 
as suggested by Fig. 119(6), and after the proportional limit 
has been exceeded, in order to keep the sum of the stresses on the 
fibers above the neutral axis equal to the sum of the stresses on the 
fibers below the neutral axis, the neutral axis must shift somewhat, 
as indicated in Fig. 119(c), thereby changing the values of c and I. 

Further, if the beam has a symmetrical section but is made of 
material, such as cast iron, whose fibers deform more in tension 
than in compression when subjected to a given unit-stress, the 
neutral axis must shift toward the compressive side of the beam in 
order to keep the sum of the tensile stresses equal to the sum of the 
compressive stressses, and this shifting of the neutral axis increases 
the resistance of the beam. 

42. Maximum Moment Due to Moving Loads.— Let several 
concentrated loads move or roll over a beam so that the distance 
between the loads remain constant, as for example, the wheel 
loads of a locomotive when moving on a bridge, and let it be re- 



Fig. 119.—Shifting of neutral axis 
in unsymmetrical section when 
proportional limit is exceeded. 













MAXIMUM MOMENT DUE TO MOVING LOADS 117 


quired to find the position of the loads when they cause the 
greatest bending moment on the beam. 

When the loads are in any given position on the beam the 
dangerous section must occur under one of the loads since the 
vertical shear will pass through zero under one of the loads, but 
although the bending moment is maximum at a section beneath 
a certain one of the loads (W 2 , say, Fig. 120) when the loads are 
in a given position, the bending moment at a section beneath 
the same load (W 2 ) after the loads have been moved to another 
position may be greater or less than it was when the loads were in 
the previous position. 

In Fig. 120 let Wi, W 2 , W 3 , W± and W 5 be concentrated loads 
that remain fixed distances 

W 3 Z, YY 

w 5 


apart as they roll over the beam 
of span l, and let it be required 
to find the position of W 2 such 
that thoben ding moment at a sec¬ 
tion beneath W 2 will be greater 
than for any other positioning. 

The resultant of the loads is 
2 IF and it will be assumed to 
act at the distance e from IF 2 . 
of equilibrium we have, 

l—x 


Wi| 


W 2 

k—a->k— b- 


a> cl) a) 


2W 
W 4 
<-d- 


dm 


l — X~( 


Fig. 120.—Moving loads. 


Ri = 


Now from one of the equations 


-2 IF. 


Let x denote the distance of W 2 from the left support, then the 
bending moment at the section beneath IF 2 is 

l—x—e 


M 3 


-xVW-Wia. 


Now the value of x which will make M x a maximum may be found 
by equating to zero the first derivative of M x with respect to x. 
Thus, 

dM x l- 2z-e vw n 

■&“ — SIF=0 - 

Hence 

l— 2 x—e_ ri 

l U ’ 


l e 

X ~2 2 ’ 


(56) 


















118 


TRANSVERSE LOADS 


and hence M 2 will be in its position of maximum moment when it is 
as far to one side of the center of the span as the resultant of all the 
loads on the span is to the other side of the center of the span. And 
the same statement will apply to each of the loads. 

In order to find the greatest bending moment to which the beam 
is subjected each load in turn must be put in its position of maxi¬ 
mum bending moment and the bending moment at the section 
beneath the load must be found; the greatest of these maximum 
moments is the greatest bending moment to which the beam is 
subjected. 

PROBLEM 

121. Two loads, 4000 lb. and 2000 lb., 6 ft. apart roll over a simple beam 
12 ft. long. Find the position of the loads to give the maximum bending 
moment and design a yellow-pine beam to carry this load, using a working 
stress of 1200 lb. per sq. in. Ans. 8 in. by 10 in. 

43. Assumptions and Limitations Involved in the Flexure 

si 

Formula. —In the derivation of the flexure formula, M =—, several 

c 

assumptions were made which impose limitations on the use of 
the formula. These assumptions and limitations may be sum¬ 
marized as follows: 

1. The unit-stress on any fiber of the beam is proportional to 
the unit-strain of the fiber, and hence the maximum fiber unit- 
stress in the beam does not exceed the proportional limit. This 
assumption of proportionality of stress and strain is a close approxi¬ 
mation to the law of behavior of most engineering materials except 
cast iron and concrete. And, for these materials the assumption 
does not as a rule introduce serious error in the flexure formula 
when the unit-stress in the beam is not greater than the usual 
working stress. 

2. The beam is composed of material for which the modulus of 
elasticity in tension is the same as that in compression. This 
assumption is a reasonably close approximation to the results of 
experiments for most engineering materials, except cast iron. 
And the error involved in the case of cast iron, although con¬ 
siderable, is not as a rule serious. The flexure formula does not, 
of course, apply directly to a beam made of two or more different 
materials such as a reinforced concrete beam. 

3. The axial strain (stretch or shortening) of any fiber in the 


LIMITATIONS INVOLVED IN FLEXURE FORMULA 119 


beam is proportional to the distance of the fiber from the neutral 
surface. This involves the further assumption that the effect of 
shearing strain on the axial strain is negligible; which, except for 
very short deep beams, introduces little error, particularly since 
the maximum longitudinal fiber stress occurs at the section on 
which the shearing stress is zero. 

4. The unit -strain as well as the total strain of any fiber of the 
beam is proportional to the distance of the fiber from the neutral 
axis. This assumption requires that all fibers shall have the same 
length before bending, that is, the beam shall be straight, and hence 
the flexure formula does not apply to curved beams. 

5. The loads act in one plane which contains the centroidal 
axis of the beam; the loads are perpendicular to the centroidal 
axis of the beam; and the neutral surface is perpendicular to the 
plane of the loads. These assumptions require that the plane of 
the loads shall contain an axis of symmetry of each cross-section 
(or, to be more exact, the plane of the loads shall contain a principal 
axis of inertia of each cross-section) in which case the neutral axis 
of any section is the other axis of symmetry (or the other principal 
axis of inertia of the cross-section). The flexure formula does not 
apply, therefore, to a beam loaded unsymmetrically. 

6. The proportions of the beam are such that the beam acts as 
a unit, that is, the beam fails (elastically) by bending and not by 
twisting, lateral collapse, local wrinkling, etc. For example, a 
rectangular beam \ in. wide by 12 in. deep would probably fail by 
twisting, and an I-beam having very wide and thin flanges would 
probably fail by local wrinkling of the flange, etc. 

Most of the steel, timber and other one-material beams that 
occur in structures and machines conform approximately to the 
conditions on which the flexure formula is based. Further, the 
flexure formula is applied frequently to beams (such as curved 
beams, unsymmetrically loaded beams, flat plates, etc.), that do 
not satisfy the conditions required to make its application valid; 
this practice is often justified if the amount of the error is known 
approximately, and if allowance is made for the error in some way 
as, for example, by reducing the working stress, or by introducing 
a correction factor in the formula. 


CHAPTER VI 


DEFLECTION OF STATICALLY DETERMINATE BEAMS 

(Double Integration Method x ) 

44. Introduction. —Beams in structures and machines when 
resisting static loads must have stiffness as well as strength; that 
is, beams must resist the loads without permitting too great an 
elastic deflection. 2 In fact stiffness is sometimes the governing 
factor in the design of the beam. 

Now the deflection of a beam depends on (1) the loads acting 
on the beam, (2) the stiffness of the material of which the beam is 
made, and (3) the dimensions of the beam. The stiffness of a 
material is measured by the modulus of elasticity of the material 
(Arts. 5 and 144). 

The purpose of this chapter is to determine in what way and 
to what extent the elastic deflection of a beam with two supports 
(statically determinate beam) depends on the loads, the stiffness 
of the material, and the dimensions of the beam, when the beam is 
loaded in various ways. To do this the general equation of the 
elastic curve of a beam will first be found. 

45. Elastic Curve Equation. —The elastic curve of a beam is the 
curve of the centroidal axis of a stressed beam, provided that 
the maximum unit-stress in the beam does not exceed the propor¬ 
tional limit of the material; since the elastic curve lies in the neutral 
surface of the beam it does not change its length as the beam is 
bent. 

The general equation of the elastic curve of a beam may be 
derived as follows: Fig. 121 represents a beam 3 bent by loads that 

1 See Chapter VIII for a discussion of the moment-area method of deter¬ 
mining deflections of beams. 

2 As discussed in Chapter XIII, the resistance of a member to impact 
loads may be decreased by making the member stiff. 

3 It is assumed that the beam is straight before the loads are applied and 
that the plane of the loads contains an axis of symmetry of each cross-section 

120 


ELASTIC CURVE EQUATION 


121 


cause a maximum fiber stress less than the proportional limit of the 
material (the deflection of the beam is exaggerated in the figure). 
As discussed in Art. 34, when a straight beam is bent the strains 
of the fibers are proportional to the distances of the fibers from the 
neutral surface. Thus in Fig. 121 GH and E'F' are any two sec¬ 
tions that were parallel before the beam was bent; the distance 
A B between the two parallel sections is assumed to be a dif¬ 
ferential length and will be denoted by dl, where l is the length of 



the elastic curve. By drawing EF through B parallel to GH 
(and hence parallel to the original position of E'F') it is evident 
that the upper fiber HF has shortened the amount FF' and the 
bottom fiber has elongated the amount EE', denoted by de in 
Fig. 121, and the strains of the other fibers are proportional to the 
distances of the fibers from the neutral surface. Further, the 
unit-strain, e, of the bottom fiber is 


de 



of the beam so that the neutral plane is perpendicular to the plane of loads 
(see Art. 43). Further, the beam is assumed to be subjected to pure bending; 
that is, the deflection resulting from shearing deformation is assumed to be 
negligible (see Art. 43). 






122 DEFLECTION OF STATICALLY DETERMINATE BEAMS 
Now, the sectors OAB and BEE' are similar, and hence 


BE = EE 
OA~AB 


But OA is the radius of curvature of the elastic curve at the 
point A, denoted by p, and BE is the distance from the neutral 
surface to the fiber whose strain is de, denoted by c as in Art. 34. 


Hence 


c de c 



(57) 


But c and e are related to the external forces, the stiffness of 
the material, and the dimensions of the beam according to the 
following equations: 


M = ~ (Art. 34) and E = S - (Art. 5). 


Therefore, equation (57) may be written: 


P = 


El 

M 


or M — 


El 


. (58) 


in which p is the radius of curvature of the elastic curve at a sec¬ 
tion for which the bending moment is ilf; jB7 is the modulus of 
elasticity of the material, and I is the moment of inertia of the 
cross-section of the beam about the neutral axis; if E is expressed 
in pounds per square inch, I in inches 4 and p in inches, M will be 
expressed in pound-inches. 

It should be noted from the above equation that if a beam of 
constant cross-section is so loaded that the bending moment M is 
constant over a portion of the beam, the radius of curvature of the 
elastic curve of this portion will also be constant (since E and I 
are constant), and hence the elastic curve for this portion is an 
arc of a circle. Conversely, if a beam is bent in an arc of a circle 
the bending moments for all sections of the beam are equal. 
Further, the above equation also shows that when M is equal 
to zero, p is equal to infinity; thus, at the inflection point (M = 0, 
Art. 38) the center of curvature is at an infinite distance from the 
beam. 


ELASTIC CURVE EQUATION 


123 


PROBLEMS 

122. A band saw 1/20 in. thick by 1 in. wide is in contact with the circum¬ 
ference of a wheel having a diameter of 50 in. Find the bending moment to 
which the saw is subjected and the unit-stress in the outer fibers. 

Ans. s = 30,000 lb. per sq. in. 

123. A steel rod is subjected to 

bending couples at its ends as shown 
in Fig. 122. If the moment of each 
couple is 2400 lb.ft. and the cross- 
section of the rod is | in. by 2 in., 
with the 2-in. dimension in the plane 
of the couple, find the radius of Fig. 122—Beam subjected to bending 
curvature of the rod. couples. 

124. Find the radius of curvature at the section AB of the beam de¬ 
scribed in Problem 73, assuming the beam to be made of yellow pine. 



Elastic Curve Equation Expressed, in Rectangular Coordinates. 
The expression for the radius of curvature, p, is (see any text¬ 
book on calculus) 



Now for beams that are straight before being bent and that 
are not deflected more than is usual in structural and machine 

dy 

members, the value of the slope is always small 4 compared 
/ dy\ 2 

with unity, and hence may be neglected without intro¬ 

ducing serious error. Thus, the above expression for p becomes 


P 


1 

d 2 y 

dx 2 


(60) 


El 

Therefore, the equation M = — may be written 

P 


M=±EI 


d\ 

dx 2 ’ 


(61) 


4 For example, the tangent to the elastic curve would probably never be 
as large as one part in 20 and hence the error in equation (60) would not 
exceed about one part in 10,000. 









124 DEFLECTION OF STATICALLY DETERMINATE BEAMS 


which is the general equation of the elastic curve of a beam. In 
this equation M is the bending moment at the section whose dis¬ 
tance from the origin of the coordinates is x, and y is the deflec¬ 
tion of the elastic curve at the same section. The sign to be 
selected for the right-hand member of the equation is discussed 
below. 

In order to determine the value of the deflection y for any given 
value of x, M is expressed in terms of x and the differential equation 
is then integrated twice; the equation thus found, of course, will 
depend on the type of beam (simple, cantilever, etc.) and the type 
of loading (concentrated, uniformly distributed, etc.). The equa¬ 
tions of the elastic curves and the maximum deflections of beams 
of various types are found in Art. 46 to 50. 

py t 

Signs of M and —In using the elastic curve equation, 


M 


(py 

±EI-~> i n this and the following chapters, it is important 

d 2 y 

to understand the significance of the signs of M and E and I 

are essentially positive and may be regarded merely as magni¬ 
tudes. Now the sign of M has already been discussed (Art. 36); 
it is positive for a horizontal beam when it produces tensile stress 
in the bottom fibers of the beam, or when it causes the center of 
curvature to lie above the beam, and is negative when it causes 


compressive stress on the bottom fibers, etc. The sign of 


d^y 

dx 2 ' 


however, depends on the choice of the positive direction of the 
Y 



Fig. 123.—Effect of direction of axes on sign of —. 

dx 2 


axes. For example, Fig. 123a represents a horizontal simple 
beam subjected to a positive bending moment. Let the origin 
of axes be chosen at the left end of the beam and let the positive 
directions for the x and y axes be to the right and upwards, 
respectively. 






DEFLECTION OF SIMPLE BEAM, UNIFORM LOAD 125 


dv 

Now the slope at a point A on the curve is negative whereas 

dv 

at a point B the slope is positive, and thus as x increases ~ 
increases. Therefore, the rate of change of with respect to x 

d 2 y\ 

that is, ^“2 ) is positive, and since M is also positive, the equation 
may be written 

.(62) 

If, however, the positive direction of the y -axis were chosen 
downward as shown in Fig. 123(6), the slope at A would be POSI¬ 
ES?/ 

tive and would decrease as x increases; thus would be negative. 

But M is positive and the right side of the equation must, then, 
also be posLi e which requires that the equation shall be written 


M = — El 


dPy 
dx 2 ’ 


(63) 


If, then, the z-axis is chosen as in Fig. 123 and the positive 
direction of the y -axis is chosen opposite to the direction of the 

d 2 y 

deflection, equation (62) should be used since the sign of w ill 

be the same as that of M, and if the positive direction of the 
2 /-axis is chosen opposite to the direction of the deflection equation 

d 2 y 

(63) should be used, since the sign of will opposite to that 
of M. 

46. Deflection of Simple Beam, Uniform Load. —Let it be 

required to determine the equation of the elastic curve and the 
maximum deflection of a simple beam, having a span of Z ft., when 
subjected to a uniformly distributed load of w lb. per ft. (Fig. 
124); the known quantities in addition to w and l are the modulus 
of elasticity E and the moment of interia, 7, of the cross-section of 
the beam. 

Let the i/-axis be chosen positive upward as shown in Fig. 124. 

d 2 y 

The general elastic curve equation, then, is F7^| = 717. But 
il7 varies with x and may be expressed in terms of x; thus, the 




126 DEFLECTION OF STATICALLY DETERMINATE BEAMS 


bending, moment at any section whose distance from the left 
support is x, is 


, _ x wl wx 2 

M x = Rix—wx-2 = ~2 X — 2 * 


Substituting this expression for M in the general elastic 
equation we have 


r^ T d 2 y wl wx 2 

EI d=2 X —2 


curve 

(64) 


By integrating we obtain, 



in which c\ is a constant of integration. The value of a constant 
in an equation may be determined by substituting a pair of values 


of the variables; in this equation the variables are the slope 
of the elastic curve and the distance x. From inspection, 


when 




By substituting this pair of values in the above equation the value 
of Ci is found. Thus 


wl 3 wl 3 
Cl=- 16“ + 48 



The above equation, then, becomes, 


. ' <*> 
































DEFLECTION OF SIMPLE BEAM, CONCENTRATED LOAD 127 


By integrating this equation and evaluating the constant of 
integration, the equation expressing the relation between y and 
x is found. Thus, 


EIy = 


wlx 3 

~w 


wx 4 
24 


wl 3 x 

"24" 


+c 2 . 


From inspection, when x = 0, y = 0, and hence c 2 = 0. 


Therefore, the equation of the elastic curve of a simple beam 
subjected to a uniformly distributed load is 


Ely 


wlx 3 wx 4 wl 3 x 
~12 15T 24"* 


( 66 ) 


Maximum Deflection .—Now the maximum deflection, A, 
occurs at the mid-span, that is, y in the above equation becomes 
A when x is equal to 1/2. Hence the maximum deflection is 


A = 

A = 


1 /wP_wl 4: wP\ 
F7V96“384 _ 48;‘ 

5 wP_ 5 Wl 3 
384 El 384 El * * 


(67) 


The minus sign shows that the deflection is opposite to the 
positive direction of the y-scx is. If the total load W is expressed 
in pounds, l in inches, E in pounds per square inch, and I in inches 4 , 


A will be expressed in inches. 

47. Deflection of Simple 
Beam, Concentrated Load at 
Mid-span. —Let the axes be 
chosen as shown in Fig. 125. 
For any section in the left half 
of the beam, 

M = R\x = \Px. 

Hence 



Fig. 125.—Deflection of simple beam 
subjected to concentrated load at 
mid-span. 


(68) 


By integrating we obtain 

Px 2 

ei £ = ~t + ci; 

when x = 1, = an d hence Ci=—jg-. 











128 DEFLECTION OF STATICALLY DETERMINATE BEAMS 


Thus, 


tjl dx 4 16 


By integrating again we obtain 


EIy = 


Px 3 PI 2 , 

12 16 X+C2 ’ 


(69) 


• (70) 


when x = 0, y = 0, and hence 5 C 2 = 0. 

Therefore, the equation of the elastic curve of the left half of 
the beam is 


Px 3 Pl 2 x 

EIy= l2- W 


(71) 


Since the maximum deflection, A, occurs at the mid-span, that 

is, when x = L the maximum deflection is 

£ 

A “ 96 16 48 EP 


The minus sign shows that the deflection is opposite to the 
positive direction of the F-axis. 

Equation (71) is also the equation of the elastic curve of the 
right half of the beam if the origin of axes is taken at the right 
end of the beam and the positive direction of x is to the left. 


PROBLEMS FOR ARTICLES 46 AND 47 

125. A simple beam 12 ft. long is subjected to a uniformly distributed 
load of 400 lb. per ft. The cross-section of the beam is 4 in. wide by 8 in. 
deep. The beam is made of oak having a modulus of elasticity of 2,000,000 
lb. per sq. in. and a proportional limit of 3500 lb. per sq. in. It is specified 
that the maximum fiber unit-stress shall not exceed 2000 lb. per sq. in. and 
that the maximum deflection shall not exceed -g-jj-Q of the span. Are the 
requirements satisfied? 

126. How much will a 12-in. 55-lb. I-beam deflect when stressed to its 

porportional limit of 40,000 lb. per sq. in. by a concentrated load at the mid¬ 
span if the span is 20 ft.? Ans. A = 1.07 in. 

5 . It should be noted that although the right support does not deflect, the 
value of y in equation (70) is not zero when x is equal to l, since 1/2 is the great¬ 
est value x can have in the equation; for values of x greater than 1/2 the expres¬ 
sion for M is not \Px. 




DEFLECTION OF CANTILEVER BEAM 


129 


127. A cylindrical steel shaft 4 in. in diameter is used as a simple beam on 

a span of 12 ft. The beam is subjected to a uniformly distributed load of 200 
lb. per ft. and a concentrated load of P lb. at the center of the span. If the 
maximum deflection of the beam must not exceed 0.40 in. and the maximum 
fiber unit-stress must not exceed 20,000 lb. per sq. in., find the maximum 
value of P. Ans. P —2290 lb. 

128. In Art. 47 use for M the bending moment for any section to the 
right of the load and show that the magnitude of the maximum deflection is 

PP 
48 El ’ 

48. Deflection of Cantilever Beam, Uniform Load.—Let the 

uniformly distributed load be w lb. per ft. and let the axes be 
chosen as in Fig. 126. 


Y 

W lb. per ft. 

X ^-rrTTTTTTI 1 1 1 1 II 1 111 


curve 

L--- 1 -> 



Fig. 126.—Deflection of cantilever beam subjected to uniform load. 


Since M is negative the elastic curve of the beam is 

W1 d 2 y __ M __wx 2 

Er &p— M - 2 . 


(72) 


By integrating and determining constants as in the preceding 
articles we have 


when x = l, ~ = 0; therefore, a = 

jpjdy wx 3 wl 3 

EI di~ ~6 6 ’ 


T7T wx* wl 3 . 
EIy=— 2 j~+-g-3+C2, 

when x = l,y = 0; therefore, C 2 = — 

The elastic curve equation then is 

pr,,_ wxi 4- wl3 x-- W U 

EI V -- 24 + 6 8 ' 


(73) 


. (74) 












130 DEFLECTION OF STATICALLY DETERMINATE BEAMS 


Now the maximum value of y in the above equation occurs when 
x is equal to 0. Hence the maximum deflection is 


1 wl 4 _ 1 wl 3 

8 EI~~8ET 


(75) 


PROBLEMS 

129. Show that the magnitude of the maximum deflection of a cantilever 
beam when subjected to a concentrated load P at its free end is 

1 PI* 

A “3 El * 

130. A 15-in. 55-lb. I-beam is used as a cantilever beam. The length of 

the cantilever is 10 ft. and the beam is subjected to a uniformly distributed 
load. Find the deflection of the free end of the beam when the maximum 
fiber unit-stress in the beam is 16,000 lb. per sq. in.; neglect the weight of the 
beam. Ans. A = 0.256 in. 

131. A piece of pine having a cross-section 2 in. wide by 4 in. deep is used 
as a cantilever beam to resist a concentrated load P at the free end. The 
beam has a free length of 4 ft. It is required that the maximum deflection 
(see Problem 129) shall not exceed 0.5 in. and that the maximum fiber unit- 
stress shall not exceed 2000 lb. per sq. in. What is the maximum value of 
P? Neglect the weight of the beam. The modulus of elasticity of the pine 
may be assumed to be 1,500,000 lb. per sq. in. 

Beam, Uniform Load.—Fig. 

127(a) represents a beam 
that overhangs the same 
amount at each end and is 
subjected t3 a uniformly- 
distributed load of w lb. per 
ft. over the entire length of 
the beam. Let it be required 
to find the equation of the 
elastic curve of the central 
portion of the beam, and the 
maximum deflection of the 
beam. 

Let the axes be chosen as shown in Fig. 127(6). The reaction 
Ri is found by applying one of the equations of equilibrium as 
follows: 

Ri — \wl-\-wa. 


49. Deflection of Overhanging 


W lb. per ft. 



AT ' 1 

. ,. Elastic curve 

( 6 ) 

Fig. 127.—Deflection of overhanging 
beam. 










DEFLECTION OF SIMPLE BEAM 


131 


The bending moment at any section between B and C (Fig. 
127a) is 

M ,= fl ia: - w(q + x)2 

= %w(a-\-l)x—\w{a+x ) 2 .(76) 

For all points between B and C: 

EI< ^= -\wx 2 +\w(l-a)x-\wa? .(77) 


By integrating this equation twice and making use of the 
following conditions, 

^| = 0 when x = y = 0 when x = 0; y = y m3jX = A when z = 

the elastic curve equation of the central portion of the beam is 
found to be 

EIy= — -^wx^+^wlx 3 —\wa 2 x 2 —^wPx+waHx; . (78) 


and the maximum deflection is found to be 

5 wl 4 3 wl 2 a 2 

“384 #T - T6 “E7~ 


(79) 


50. Deflection of Simple Beam, Concentrated Load Not at 
Mid-span.—If the concentrated load P is not at the mid-span the 
expressions for M on opposite sides of the load are different as is 
also the case when the load acts at the mid-span (see Art. 47), and 
hence there are two elastic curves which have different equations. 
But, if the load does not act at the mid-span the constants of inte¬ 
gration cannot be determined as in Art. 47; they are found by 
making use of the fact that the two elastic curves have a common 
tangent and a common 
ordinate under the load. 

Further, the maximum 
deflection of the beam is q 
the maximum value of 
y in the equation of only 
one of the curves. 

Fig. 128 represents p IG 128.— Deflection of simple beam; concen- 
a simple beam subject- trated load at any point, 

ed to a concentrated 

load P at the distance a from the left support and b from the right 
support, a being greater than b. Let it be required to find the 














132 DEFLECTION OF STATICALLY DETERMINATE BEAMS 


equation of the elastic curve of the left portion of the beam and the 
maximum deflection of the beam. Let the axes be chosen as 
shown in Fig. 128. 

The bending moment at any section to the left of P is 


M = R\X= -yX, .(80) 

in which x may have any value from 0 to a. And the moment 
at any section to the right of P is 

M = ^x—P(x — a), . (81) 


in which x can not be less than a nor greater than l. 

For all points to the left For all points to the right of the 
of the load: load: 


VT d 2 y Pbx 

EI dx^ = ~r ■ ■ 

dy_Pbx 2 

EI di~^T +Cl - 


(82) 


(84) 


E1 %=T x - p ^~ a) - 


(83) 


„ jdy Pb 2 P{x—a) 2 , 

EI Tx=2l X -—' (85) 

And since the curves have a common tangent under the load, 

dy 

when x is made equal to a in both (84) and (85) the value of ~ 

in (84) is equal to in (85). Thus 

Pba 2 , Pba 2 P(a—a) 2 . 

~W +Cl = ^l - 2 - +C3 ’ 

whence, 

Ci = C 3 . 

Now by substituting Ci for c 3 in (85) and by integrating (84) 
and (85) there is found: 

Phr 3 

EIy = —Qj —bci*+c 2 . (86) 


Pbx 3 P{x—a) 3 

Ely —^--g-bcix+c 4 . (87) 


when x = 0, y = 0, 
hence, C 2 = 0. 

Since the curves have a common ordinate under the load, when 
x = a in (86) and (87) the values of y in these equations are equal, 
and hence 

Pba 3 . Pba 3 


6 1 


-+cia = - 


6 1 


■+Cia+C4. 


C4 = 0. 















PROBLEMS 


133 


Now when x = l in (87), y = 0, and hence 

Pbl 2 , P(l-a)* = _Pb 
6 1 


~ « 2 ~b 2 ). 


( 88 ) 


6 1 1 6Z 

Substituting this value of c\ in (86) the equation of the elastic 
curve of the left portion of the beam is found to be 
Pbx 3 Pb(l 2 —b 2 )x 


Ely 


6 1 


6Z 

dy 


(89) 


The value of x that makes ^ equal to zero is the value of x 

that makes y in (89) a maximum, and this value of y is the value 

of the maximum deflection. But the value of ^ is given by (84) 

ax 

and hence equating (84) to zero we have 

2 _ l 2 — b 2 _a(a+26) 

* 3 3 ' 

By substituting this value of x in (89) the maximum deflection 
is found to be 

Pb(l 2 -b 2 )Vd(l 2 -b 2 ) 


A= — 


A= — 


27 Ell 

Pba{a-\-2b) V3 a (a +26) 
27 Wl 


(90) 


( 91 ) 


If in the above expression a and b are each made equal to 


1 PI 3 

that is, if P acts at the mid-span the value of A is -rpr as was 
’ ^ 48 El 

found in Art. 47. 

PROBLEMS 

132 . Derive the equations of the elastic curves of a simple beam of length 
l when loaded with two equal concentrated loads, PP, each load being at a 
distance of \l from a support. Also find the maximum deflection of the beam. 
Select the origin of axes at the left end of the beam and the positive direction 
of the x and y axes to the right and upwards, respectively. 

Px 3 3 

Ely —— -— Pl 2 x for the left portion, 

6 32 

Plx 2 Plx PI 3 

Ans. Ely =—--f° r central portion, 

8 8 o84 


A= - 


U_ PP 
384 El' 


Note. —The elastic curves of the left and central portions of the beam have 
a common tangent and a common ordinate under the left load. 















134 DEFLECTION OF STATICALLY DETERMINATE BEAMS 


133 . A 9-in. 30-lb. I-beam rests on two supports 14 ft. apart and resists a 
concentrated load of 15,000 lb. at a distance of 5 ft. from the left support. 
Determine whether the following requirements are satisfied. 

(а) Fiber unit-stress not to exceed 16,000 lb. per sq. in. 

(б) Maximum deflect not to exceed g-g-g of the span. 


Deflection of Beam Due to Shear 
The deflection of a beam due to the shearing stresses in the 
beam was assumed, in the preceding discussions, to be negligible. 

In short deep beams, however, the 
deflection due to shear may require 
consideration. An approximate value 
for the deflection due to the shear¬ 
ing stresses may be found as fol¬ 
lows: 

The shearing deformation dy s of 
a small block, of length dx, in the 
beam (Fig. 129) is 



fl dx 



BpW 

Ss 

t 1 


Fig. 129.—Deflection of beam 
due to shear. 


dy s = e s dx = ~dx, 

in which s s is the shearing unit-stress on the face of the block; 
the deflection of a fiber of length x (Fig. 129) is 

.(92) 


ys = 4rf s 8 dx. 


If s s is constant throughout the length of all the fibers the deflec¬ 
tion of the beam is 

s R x 


y ‘ e: 


(93) 


In general, however, varies over the section, but if an average 
value of s s (see Art. 40) is used in equation (93), an approximate 
value of the deflection due to shear may be found which may be 
helpful in estimating to what extent the shearing stresses con¬ 
tribute to the deflection of a beam. 

A more exact analysis may be made by substituting for s s 
in equation (92) the value given by equation 53 of Art. 40. 

PROBLEM 

134 . Since Vdx = dM x (see Arts. 36 and 37) and the average shearing unit- 
stress s s =V/a show that equation (93) may be written 

M x 
Vs ~ aEg 

when the vertical shear, V, is constant over the length x. 







CHAPTER VII 


STATICALLY INDETERMINATE BEAMS 


51. Introduction.—In the preceding two chapters the beams 
considered were held in quilibrium by external forces that formed 
a system of parallel forces in a plane, and the reactions of the sup¬ 
ports were found by applying the two equations of equilibrium for 
such a force system; since there were not more than two supports 
and hence not more than two unknown external forces acting on 
the beam, the two equations of equilibrium were sufficient to deter¬ 
mine all the unknown external forces. In other words, the force 
system acting on the simple, cantilever, and overhanging beams 
considered in the preceding chapters were statically determinate. 

On the other hand, although fixed-ended beams and continuous 
beams are held in equilibrium by parallel forces in a plane, yet the 
reactions of the supports cannot be found from the two equations 
of equilibrium alone, since the number of unknown reactions is 
greater than two. Such beams are said to be statically indeter¬ 
minate, and in finding the reactions use is made of the relation 
between the external forces and the internal effects of those forces 
(as expressed by the elastic curve equation) in addition to the 
relations between the external forces alone (as expressed by the 
equations of equilibrium). 

Method of Procedure. —The method of obtaining equations 
that involve the unknown reactions, in addition to the two equa¬ 
tions of equilibrium, is as follows: 


In the elastic curve equation, M=±EI 


d?y 
dx 2’ 


the expression 


for M will contain some of the unknown reactions (forces and 
couples) acting on the beam, and after the elastic curve equation 
is integrated, a constant of integration being added with each 
integration, the unknown reactions and constants of integration 
are determined by substituting in the equations as many pairs of 
values of the variables as there are unknown reactions and con- 


135 


136 


STATICALLY INDETERMINATE BEAMS 


stants of integration; the pairs of values of the variables are found 
from the physical conditions that the beam satisfies. The remain¬ 
ing unknown forces (that did not occur in the expression for M) 
may then, as a rule, be found from the two equations of equi¬ 
librium. 


Thus, the equation of the elastic curve may be used for two 
different purposes: (1) To find the deflection of a beam when all 

the forces acting on the beam 
are known, as discussed in 
Chapter VI, and (2) to deter¬ 
mine reactions of supports 
when they cannot be found 
from the equations of equilib¬ 
rium alone, as is discussed 
in this chapter in connection 
with fixed-ended and contin¬ 
uous beams; the use of the 
elastic curve equation for the 
latter purpose is of great im¬ 
portance, since it makes pos¬ 
sible the determination of the 
stresses in statically indetermi¬ 
nate flexural members. 

52. Beam Fixed at One 
End, Supported at Other End; 
Uniform Load. —Fig. 130(a) 
represents such a beam, the 
tangent line to the elastic 
curve at the wall being hori¬ 
zontal; the length of span is 
l, and the load per unit of 
length is w. As shown in 
Fig. 130(5), the beam is held 
in equilibrium by parallel 
forces of which all three reac¬ 
tions, Ri, R 2 and R 3 are un¬ 
known. Thus it is necessary, as explained in the preceding 
article, to use the elastic curve equation in addition to the two 
equations of equilibrium in order to determine the reactions. 

Now the forces R 2 and R 3 are equivalent to an upward force 



Fig. 130/—Beam fixed at one end, sup¬ 
ported at other end; load uniformly 
distributed. 
















































BEAM FIXED AT ONE END 


137 


and a bending couple, and since we are concerned mainly with the 
part of the beam outside the wall, it is convenient to assume that 
the beam is cut off even with the face of the wall at the section A A 
(Fig. 1306) and that R 2 and Rs are replaced by the forces 
(stresses) which they cause on this section; namely, the upward 
shearing force Fo and the bending couple Mo as shown in 
Fig. 130(c). 

The three reactions to be found, then, are Ri, Vo and M 0 . 
The value of Ah may be found from the elastic curve equation, and 
Vo and Mo may then be found from the two equations of equilib¬ 
rium as follows: 

If axes are chosen as shown in Fig. 130(c) the elastic curve 
equation (see Art. 45) is 

EIj£ i = M = R l x--~ .(94) 


Integrating, we have 


WT dy Rix 2 wx 3 

m Tx=—--<T +Cu 


dy 


when x = l, ~r = 0 and hence, 
ax 


Therefore, 


Ril 2 . wl 3 

-njdy _R\x 2 wx 3 Ril 2 wl 3 
M di~ 2~ ~2 ^~6~' 


Integrating again, we obtain, 


„ T Rix 3 wx 4 R\l 2 wl 3 2 

EI V = -q— 2i—2-*+T* +A • 


(95) 


(96) 


when x = 0, y = 0 and hence, C 2 = 0; 
when x = l, y = 0 and hence, 

Rxl 3 wl 4 Ril 3 wl 4 
° 6 24 2 + 6 ’ 

Therefore, 

Ri = %wl. 

Now from the two equations of equilibrium (2F = 0 and 2ikf = 0), 
Vo and Mo may be found. Thus, 

2F = %wl+Vo—wl = Q. 






138 


STATICALLY INDETERMINATE BEAMS 


Hence, 

Vo = f wl 

Hence, 

Mo= — \wl 2 . 

It will be noted that the moment at the wall, then, is a negative 
moment, as shown in Fig. 130(c), the magnitude of which is \wl 2 . 

Shear and Moment Diagrams. Maximum Positive Moment .— 
The vertical shear for a section at distance x from the left support 
is 

V x = i wl—wx, 

and this equation shows that the shear is zero when x = f l; the 
shear diagram is shown in Fig. 130(d). 

The bending moment for a section at the distance x from the 
left support is 

M x = lwlx—wx . (97) 


The value of the bending moment is a maximum at the section 
for which the vertical shear is zero (Art. 35), and hence the maxi¬ 
mum positive bending moment is 


Pos. M max = %wl-U-w-%l-i 1 1 


xfs^ 2 * 


The inflexion point is found by equating M x to zero. Thus, 


whence, 


3 7 WX n 

f wlx — 2 ~ = 0, 
x = U. 


The bending moment diagram is shown in Fig. 130(d). 

Since the maximum negative moment { — \wl 2 ), which occurs 
at the wall, is greater than the maximum positive moment, the 
greatest unit-stress in this type of beam occurs at the wall and 


may be found from the flexure formula (m~— ) by using \wl 2 
for M. ' c ' 

Maximum Deflection .—Equation (96) is the equation of the 
elastic curve, the value of C 2 being zero and the value of R i being 
| wl. Thus, the elastic curve equation becomes, 


24 Ely = %wl(x 3 — 3 1 2 ) —w(x*—41 3 x ). 



BEAM FIXED AT ONE END 


139 


Further, the maximum deflection is found to occur at the distance 
x = 0.42152 from the left support and its value is 

711/4 

A = 0.0054^. 


As indicated in Fig. 130(e), the beam may be considered to be a 
combination of a simple beam having a span J l and a cantilever of 
length J l. That is, the beam may be sawed in two at the section 
where the moment is zero (inflection point), the shear V' at this 
section acting as the reaction for the simple beam and as a con¬ 
centrated load on the cantilever. 

Alternative Method .—If the left support were removed the beam 
would become a cantilever beam and the uniform load on the 

1 wP 

beam would cause the free end to deflect a distance 3 ^ (Art. 48). 

o El 

But since the end does not deflect, the reaction R\ must be a force 
which if acting alone on the end of the cantilever would cause an 


upward deflection equal to 

O El 


Now the deflection due to a. 


1 PI 3 

concentrated load P on the free end of a cantilever is ^ 

6 hi 


(Prob. 129). Hence, 


Therefore, 


1 Ril 3 
3 El 


1 wP 

8 W 


Ri = Iwl. 


This method does not, however, avoid the use of the elastic 
curve equation since the elastic curve equation was used in deter¬ 
mining the expressions used for the deflections. 


PROBLEMS 

134 . Derive the expressions for the maximum positive moment, the 
maximum negative moment, and the maximum deflection, as given above, 
by considering the beam to be composed of two beams as shown in Fig. 130(e). 

135 . A steel I-beam is fixed at one end and supported at the other end, the 
length of span being 10 ft. It is loaded with a uniformly distributed load of 
100 lb. per ft. Find (a) the point of inflection, (6) the maximum negative 
moment, (c) the maximum positive moment, ( d ) the section modulus of the 
beam using a working unit-stress of 16,000 lb. per sq. in. Also draw to scale- 
the moment and shear diagrams. 

Ans. (d) section modulus = 0.937 in. 3 if weight of beam is neglected. 


140 


STATICALLY INDETERMINATE BEAMS 


136. A timber beam 6 in. wide and 12 in. deep is used on a span of 16 ft. 
The beam is fixed at one end and supported at the other and is subjected to a 
uniformly distributed load. If the working unit-stress is 800 lb. per sq. in., 
what is the maximum load per foot of length the beam can resist. 

137. A 10-in. 25-lb. I-beam is supported at the left end and partially 
restrained in a wall at the right end, the clear span being 15 ft. The load 
including the weight of the beam is 1200 lb. per ft., and the reaction at the left 
end is wl. Find the fiber unit-stresses in the beam due to the negative and 
positive bending moments and indicate on a sketch where these stresses occur. 


53. Beam Fixed at Both Ends; Uniform Load.—Fig. 131(a) 
represents such a beam, the length of span being l and the load 



Fig. 131.—Beam fixed at both ends; 
load uniformly distributed. 


per unit of length, w. Fig. 
131(h) shows that the beam 
is held in equilibrium by a 
parallel force system in which 
there are four reactions, Ri, 
R 2 , R 3 and R±. But, as in 
the preceding article, it is 
desirable to replace these 
reactions by the shears and 
moments at the ends of the 
beams as indicated in Fig. 
131(c), and hence the four 
unknown reactions are V\, 
V2, Mi and M2. Now from 
symmetry 1 and from the 
equations of equilibrium we 
have: 

Mi=M2 and Vi = V 2 =iwl, 

and hence only one unknown 
(namely, M 1 or M 2 ) need 
be found from the elastic 
curve equation. The value 
of Mi may be found as fol¬ 
lows: 


x If the condition of symmetry were not employed, two unknowns ( 7 i 
and Mi) would be found from the elastic curve equation, since only two equa¬ 
tions of equilibrium are available. 




















































BEAM FIXED AT BOTH ENDS 


141 


If axes are chosen as shown in Fig. 131(c) the elastic curve equa¬ 
tion is (see Art. 45) 

EI^ = M = M 1 + V 1 x-^-- ..... ( 98 ) 

Mi will be considered to be unknown both in sign and in magni¬ 
tude. Integrating, we have 

cLy 

^ = 0 when x = 0, hence C\ = 0, 

~ = 0 when x = l, hence 
ax 

n i\/r 7 I i r ^ 

0 —^-g-. 

Therefore, 

Mx=-^wP. 

Thus the moment at each wall is negative and equal to -fawl 2 . 

Shear and Moment Diagrams. Maximum Positive Moment .— 
The vertical shear for a section at the distance x from the left 
support is 

V x =iwl—wx; 

this is equal to zero when x = ^l, that is, at the mid-span. The 
shear diagram is shown in Fig. 131(d). 

The bending moment about a section at a distance x from the 
left support is 

. M x =Mi + Vix-~ 

= -^wP+iwlx- W £. 

But the bending moment is maximum at the section for which the 
vertical shear is zero and hence, 

l wl 2 

Maximum Positive Moment = —-^wl 2 —~- 

Z o 

1 79 




142 


STATICALLY INDETERMINATE BEAMS 


Thus, the bending moment at the center of the beam is only one- 
half as large as that at the ends of the beam. 

The points of inflection are found by equating the expression 
for the bending moment, M x , to zero. Thus 

M x = —&wl 2 +hwlx-^- = 0, 

whence 

z = iZ(l±|V3). 

The bending moment diagram is shown in Fig. 131(d). 

Maximum Deflection. —The expression for the maximum 
deflection may be found from the elastic curve equation; if the 
axes are chosen as in Fig. 131(c), the elastic curve equation is, 

EIjx i= M X = -&wP+hdx--£. 

By integrating this equation twice, determining constants of 
integration, etc., the value of the maximum deflection is found to 
be, 

A — — * w ^ 

384 ffl’ 

which is only one-fifth as large as that of a similarly loaded simple 
beam (see Art. 46). 

It is sometimes convenient to consider the beam in question to 
be made up of two cantilevers and a simple beam as indicated in 
Fig. 131(e). The shears V' at the points of inflection are 

V' = Vi-wx 

= \wl —— }a/3) . 

PROBLEMS 

138 . Select a steel I-beam for a span of 12 ft. if the beam is to be fixed at 
both ends and subjected to a uniformly distributed load of 200 lb. per ft. 
Use a working unit-stress of 18,000 lb. per sq. in. 

Arts. 3-in. 5Hb. I-beam. 

139 . Compare the maximum deflections of a timber beam having both 
ends fixed and a steel beam having simply supported ends. The beams are 


BEAM FIXED AT BOTH ENDS 


143 


subjected to the same uniformly distributed load, and have the same span and 
dimensions. 


140 . A beam with a clear span of 16 ft. is partially fixed at each end, the 

negative moment at each end being wl 2 - The load is uniformly distrib¬ 

uted and equal to 1000 lb. per ft., including the weight of the beam. Find the 
size of the lightest steel I-beam that may be used without having the fiber 
unit-stress exceed 16,000 lb. per sq. in. 

141 . One of the beams in a theater balcony is 18 feet long. The left end 
is built into the wall in such a manner that the restraint produces a negative 
bending moment, M 0 , at that end equal to 12,000 ft.-lb., and the beam rests 
on a support 3 feet from the right end. The beam is loaded with a uniformly 
distributed load of 600 lb. per ft. Determine the reactions on the beam, con¬ 
struct a shear diagram, and determine the value of the maximum positive 
bending moment. 


54. Beam Fixed at Both Ends; Concentrated Load at Mid- 
span. —Fig. 132(a) represents such a beam, the length of span 
being l and the load P. The 
reactions of the wall may be 
replaced by the shears V i and 
V 2 and the moments Mi and 
M 2 (Fig. 1326) as was done 
in the preceding articles. 

These four unknown reactions 
may be found by use of the 
conditions of symmetry and 
equilibrium, and of the equa¬ 
tion of the elastic curve of 
the left half of the beam. 

Thus, from the ‘ conditions of 
symmetry and equilibrium we 
have, 

Mi=M 2 and Vi = V 2 = JP. 



,-Shear curve 


& 

^Jr 

s 


urve ^ \ " As 


K-44 

P= 




P 


,Vswl 

-Moment 

curve 




(<0 


-Sp- § 

£ 


Fig. 132.—Beam fixed at both ends: 
concentrated load at mid-span. 


Now Mi may be found from 

the elastic curve equation; if axes are chosen as shown in Fig. 
132(6) the equation of the elastic curve of the left half of the 
beam is 


El 


d?y 

dx 2 


= M 1 +Vix=M 1 +iPx. . . 


. (99) 




































144 


STATICALLY INDETERMINATE BEAMS 


Integrating, we have 


EI i-M„+^+ a 


(100) 




dx 2 



Therefore 


M >—5«- 


Thus the bending moment at each end of the beam is negative and 
equal to \Pl. 

The maximum positive moment occurs at the center where the 
vertical shear is zero (see Fig. 132c for shear diagram) and is 



Thus the maximum positive moment is numerically equal to the 
maximum negative moment, and hence the maximum fiber unit- 
stresses at the wall and mid-span sections are equal, if the beam 
has a constant cross-section. 

The inflection point in the curve of the left half of the beam 
is found, by equating the expression for the bending moment 
( — |PZ+ %Px) to zero, to be half way between the wall and the 
center of the span (z = JZ). The bending moment diagram is 
shown in Fig. 132(c). 

By integrating equation (100) and observing that y = 0 when 


£ = 0, and that y = A when x=-, the maximum deflection is found 
to be 2 


1 PP 
192 EP 


which is only one-fourth as much as that of a simply supported 
beam similarly loaded (Art. 47). 



BEAM FIXED AT ONE END 


145 


PROBLEMS 

142 . A timber beam 6 in. wide and 12 in. deep is fixed at both ends and 

has a span of 16 ft. If it is subjected to a concentrated load at the center, 
what load can it safely resist if a working unit-stress of 800 lb. per sq. in. is 
used? Ans. P — 4800 lb. 

143 . Find the maximum deflection of the beam described in the preceding 
problem when the beam is subjected to a maximum fiber unit-stress of 800 
lb. per sq. in. 


l c * J 

jP l 

r 

\ m 

1_ [ 

7$b 

b ; 

p («> m. 

• 


55. Beam Fixed at One End Supported at Other End, Concen¬ 
trated Load at Mid-span.— Fig. 133(a) represents such a beam, the 
length of the span being l and the 
load, P. As in the preceding articles, 
the reactions exerted by the wall 
may be replaced by the shear V 2 
and the moment M 2 acting on the 
section at the wall as shown in Fig. 

133(6). Thus there are three un¬ 
known reactions (Ri, V 2 ,M 2 ) to be 
found. The left reaction Ri may be 
found by use of the equations of the 
two elastic curves; the expressions 
for the bending moments on oppo¬ 
site sides of the load are different, 
and hence the elastic curves on op¬ 
posite sides of the load are differ¬ 
ent; but the two curves have a 
common tangent and a common 
ordinate under the load. After Ri 
has been found from the elastic 
curve equations, the two equations 
of equilibrium may be used to 
find the values of V 2 and M 2 . 

The procedure is as follows: Let the axes be chosen as shown in 
Fig. 133(c). The equations for the elastic curve of the left half 
of the beam are: 



Fig. 133.—Beam fixed at one end, 
supported at other end: concen¬ 
trated load at mid-span. 


EI j^=Rix .( 101 ) 

EI^=lRix 2 +ci .( 102 ) 

ax 

EIy = lR 1 x 3 +c 1 x+c 3 . (103) 







































146 STATICALLY INDETERMINATE BEAMS 

The similar equations of the elastic curve of the right half 


are: 

EIj^=Rix-P(x-\l) .(104) 

El'^t =ift ix 2 - Px 2 +iPlx+c 2 .(105) 

EIy = iRix a -lPx 3 +lPlx 2 +c 2 x+c i . . . (106) 


The five unknowns (ci, C 3 , c 2 , C 4 and Pi) may be found by 
making use of the following conditions: In (103) y = 0 when 

x = 0 ; in (105) j| = 0 when x = J; ^ in (102) = in (105), when 

x = ^ in both (102) and (105); y in (103) =y in (106), when x ~2 

in both (103) and (106); and, in (106) y = 0 when x-l. 

After the values of the constants are substituted in (103) and 
(106) the following equations of the elastic curves are found. 

24JEIy=4R 1 x 3 +3Pl 2 x-12Ril 2 x .(103a) 

48 Ely = 8 Pix 3 - 8 Pz 3 +12 Plx 2 - 2AR 1 l 2 x+Pl 3 , . (106a) 

and by using the last of the conditions stated above, namely, 
y = 0 in (106) when x — l, the value of R\ is found to be 

Now from the two equations of equilibrium we have (see Fig. 
1336), 

2 F = ^P+7 2 -P = 0 . 

Hence, 

v 2 

ZM B = £Pl-Pi-M 2 = 0. 

Hence, 

m 2 =—&pl 

The maximum positive moment occurs at the mid-span and is, 
Maximum Positive Moment = y&P X 2 1 — tiPI- 

The shear and moment diagrams are shown in Fig. 133(d). 

56. Comparison with Simple Beams. —In the preceding articles 
it has been shown that a fixed-ended beam is stronger and stiffer 
than a similar beam that is simply supported at its ends. This 
is true because the positive bending moment in the central portion 
of a beam is decreased by the amount of the negative moment at 




COMPARISON WITH SIMPLE BEAMS 


147 


VJ lb. Der ft. 


the ends, and thus more of the material of a fixed beam is effective 
in resisting the loads than in the simply supported beam, since the 
bending moment is distributed more nearly uniformly along the 
length of the fixed beam. 

The effect of applying negative moments at the ends of a 
simply supported beam is shown in Fig. 134. The curve CED, 
referred to CD as a base line, represents the bending moment 
diagram for a simply supported beam subjected fco a uniformly 
distributed load of w lb. per ft., the maximum bending moment 
(as shown in Art. 36, Prob. 88) is at the mid-span and is equal to 
\wl 2 . Now, if the beam were a fixed-ended beam the negative 
moments at the ends would be 
■fawl 2 (as shown in Art. 53) 
and the curve CED would still 
represent the moment diagram 
if AB, Fig. 134(6), instead of 
CD, represents the axis or base 
line. Thus the moment at the 
center is decreased by the 
amount of the negative moment 
at the end, and further, the sum 
of the maximum positive and 
negative moments in a beam 
fixed at both ends is equal to the 
maximum moment in a simi¬ 
larly loaded simple beam. The student should show that this 
statement is also true if the load is concentrated at the mid-span. 

In practice, however, it frequently is difficult to determine to 
what extent a beam is restrained at the ends, since yielding of the 
end-restraining bodies such as abutments, riveted end-connec¬ 
tions, etc., decreases the negative moments at the ends and 
increases the positive moment by an equal amount. Likewise, 
unequal settlements of the ends and temperature changes influ¬ 
ence the stresses in fixed-ended beams to a greater extent than in 
simply supported beams. For these reasons fixed-ended beams 
are not used in preference to simple beams as extensively as the 
above results would seem to justify. However, test results 2 
obtained with certain types of riveted end-connections show that 

2 Tests to Determine the Rigidity of Riveted Joints of Steel Structures. 
Bull. No. 104, Engineering Experiment Station, University of Illinois. 



(f>) ° 

Fig. 134.—Relation between moment 
diagrams for fixed and simple beams. 






















148 


STATICALLY INDETERMINATE BEAMS 


the moments at the ends are reduced but little by the yielding of 
these riveted end-restraining connections. 

57. Continuous Beams. Theorem of Three Moments.—A 
continuous beam is one that rests on more than two supports. 
Since the .number of reactions is greater than two, use is made of 
the elastic curve equation in order to obtain the reactions of the 
supports, the two equations of equilibrium alone being insufficient. 

However, instead of determining the reactions first and then 
finding the bending moment (and hence fiber unit-stress) at any 
section, it is more convenient to determine first the negative 
moments over all the supports and then to find the reactions of the 
supports and the bending moments and vertical shears at other 
sections from these negative moments. The negative moments 
may be found by use of the theorem of three moments which may 
be derived from the elastic curve equations of two adjacent spans 
as follows: 

Theorem of Three Moments, Load Uniformly Distributed .— 
Fig. 135(a) represents a continuous beam of several unequal spans; 


w 4 

imi'Mimm 


m 


L -l —u 


1 

T 



p v l C/ 2 

Y W 2 lb. per ft. | 

(a) 

" 3 

—T~ 

1 

i 4 



(&) V. £ 

M " 


>C 

fat 


V, 


i 

i 

W s lb. per ft. I 


£ 


(c) 


Fig. 135. Moments at supports of continuous beam. 


each span is subjected to a uniformly distributed load, but the 
load is not constant over the whole length of the beam; all 
supports are assumed to be on the same level. Consider'two 
adjacent spans (second and third spans) in Fig. 135. Let M 2> 
Ms, M 4 , etc., denote the bending moments at the second, third' 
fourth, etc., supports. Let V 2 denote the shear on a section just 
to the right of the second support and V- 2 just to the left of the 
second support; similarly let F 3 and F - 3 denote the shears on 
the right and left, respectively, of the third support, etc. 

A free-body diagram showing all the forces acting on that 
part of the beam between the second and third supports is shown 























CONTINUOUS BEAMS 


149 


in Fig. 135(6) and a similar diagram for the part of the beam 
between the third and fourth supports is shown in Fig. 135(c). 

For the second span let the axes be chosen as indicated in 
Fig. 135(6). Then, the elastic curve equation for the second span 

EI^y = M 2 +V 2 x-^f 2 -. . ..(107) 

EI t= M > x+Y ¥-nr +Cl .< 108 > 

M 2 X 2 , V 2 X 3 W2X 4: . , 

EIy=- - 2 T+cix+c 2 . . . . (109) 

Similarly for the third span let the origin be chosen at the 
second support as shown in Fig. 135 (c). Then 

EI^Ms+Vsx-^ .(110) 

EI d £=M 3X +Yf -^ +c3 . (111 ) 

EIy=^f+Vf-^+c 3*+ C4 . . . . (112) 

Now in (109), y — 0 when x = 0 , and hence C2 = 0; also y — 0 

when % = 2 and hence the value of ci may easily be obtained. 

Again, in (112) y = 0 when x = 0, and hence C 4 = 0; also y = 0 
when x = h and hence the value of cs may be obtained. Further 

^ in (108) is equal to ^ in (111) when x in (108) is equal to h 

and x in (111) is equal to zero. By making use of these facts 
the following equation is found. 

12M 2 l2-\-SV2l2 2 -Sw 2 l2 3 =-12M 3 l 3 -4:Vsls 2 +WBh 3 . . (113) 

Now F 2 and V 3 may be expressed in terms of M 2 , M 3 , M± 
and the known quantities h , l 3 , W 2 and w 3 by using one of the 
equations of equilibrium for the forces in Fig. 135(6) and in 
Fig. 135(c). Thus, for the second span (Fig. 1356) one equilibrium 
equation is 

2MC = M 2 +V 2 I 2 ~ hw 2 l 2 2 +M 3 = 0 . 

_ — M2+\w2h 2 — M 3 


Hence, 


(114) 









150 


STATICALLY INDETERMINATE BEAMS 


and for the third span, 

ZM D = Ms+V3l-iw 3 h 2 +M4, = 0. | 

Hence, /i i k\ 

Tr -M 3 +iw 3 h*-M, [ ' * ‘ * Ui5; 

K3_ l J 

in which all the moments are assumed to be positive. 

By substituting these values of V 2 and V 3 in equation (113) 
the following relation between the bending moments at three 
consecutive supports is found: 

M 2 l 2 -\-2M 3 (l 2 -\~l 3 ) -\-M±l 3 = — \w 2 Z 2 3 i^3^3 3 j 

or, if the subscripts 1, 2, and 3 are used to refer to any three con¬ 
secutive supports and spans the equation may be written, 

M 1 h+2M 2 (h+l2)+M 3 l 2 = -iwiZi 3 -iw 2 fe 3 . . (116) 

which expresses the theorem of three moments for a continuous beam 
subjected to uniformly distributed loads and resting on supports 
on the same level. 

If the spans are equal (li=h = l) and each span carries the 
same load (w 2 = w 3 = w) the equation becomes 

Mi+ 4M 2 -\-M 3 = —%wl 2 , .... (117) 

which expresses the theorem of three moments for a continuous 
beam with equal spans and with a constant uniformly distributed 
load over the entire beam. 

58. Solution of Typical Problem. —Let it be required to draw 
the shear and moment diagrams for a continuous beam of four 
spans, each of length l, when subjected to a load of w lb. per ft. 
over the entire beam, and also to find the maximum fiber unit- 
stress and the equation of the elastic curve of the first span. 
(See Fig. 136(a).) The known quantities are: w, l, E and I. The 
main steps in the solution are: 

(a) By use of the theorem of three moments find the 
negative bending moment over all supports. 

(h) Find the reactions of the supports from the negative 
moments over the supports; all the external forces acting on 
the beam will then be known. 

(c) Find the vertical shears at various sections and draw 
the shear diagram. 



SOLUTION OF TYPICAL PROBLEM 


151 


(d) Find the bending moments at various sections and 
draw the moment diagram. 

si 

(e) Equate the maximum bending moment to — and solve 

c 

for s, the maximum fiber unit-stress. 


(/) In the elastic curve equation EI-~ = M substitute 

for M the expression for the bending moment in the first span 
and integrate twice, etc. 



Fig. 136.—Moment and shear diagrams for continuous beam. 


Solution. —(a) In accordance with the theorem of three moments, the fol¬ 
lowing equations may be written: 

M 1 +4M 2 +M 3 =-iwl 2 , 

M 2 +±Mz-\-M 4 =* 

Af 3 +4Af 4 +Af 5 = - \wlK 

Further, M 1 = M h = 0 since there is no restraint at the ends; and M 2 = M 4 , 
from the condition of symmetry. The solution of these equations gives the 
following values: 

Mi = 0, M 2 =~z\wl*, Mz = —£gwl\ Mi = --%-qwI 2 , M 5 = 0. 

(6) Now the bending moment at a section over the second support as 
found above is —-^gwl 2 , but the bending moment at any section is the 
algebraic sum of the moments of the forces to the left of the section. 
Hence, 

R\l — wl ~=—j Ri = -^-j^wl. 

A 


























































152 


STATICALLY INDETERMINATE BEAMS 


Similarly for the moment over the third support we have, 

R i • 21R 2 I — w-2l-l = — ~^qwI' } R 2 —^^wI. 

In a similar way the value of R 3 is found to be, R 3 =%%wl, and from sym¬ 
metry, R a = R 2 and R 5 = Ri. 

(c) The shear and moment diagrams are shown in Fig. 136(6) and (c). 
The vertical shear at any section in the first span at the distance x from the 
left support is 

V x — R\—wx = ^ wl—wx , 


and hence the vertical shear is zero when x=^l; the maximum positive 
moment occurs at this section. 

The shear just to the left of the second support is 

V — 2 = R 1 — Wl — 7 £-^wI — wl — — -g-g wlj 


and the shear just to the right of the second support is 

V 2 = R i~\~ Ra — wl—yji^wl T - ij~g -wl—wl = T" -wl. 

Thus, the shear changes at the second support from wl to +^wl 
due to the reaction R 2 ’, in other words, the reaction of a support is equal 
to the arithmetic sum of the shears at the two sides of the support. 

The shears in other spans may be found by the same method as used 
above for the first span; the values are given in Fig. 136(6). 

Thus the reaction at any support may be found by solving for the shears 
on the two sides of the support and adding them rather than by the method 
used under (6) above. 

( d ) The bending moment at any section in the first span is 


M' x = R lX - 


wx 2 
~ 2 } 


and as noted above this is maximum when x = ^l. Therefore, the maximum 
positive moment is 


11 7 11 w/llA 2 121 

Max. pos. moment = —wl • —l — ( —l ) =- wl 2 , 

P 28 28 2\28 / 1568 


which is less than the maximum negative moment (-^ %wl 2 ). 
The inflection point occurs where M' x = 0; thus 


whence 


11 _ wx 2 

M *=M wl ' x ~T =0 ’ 


z 2 8 l 


The bending moment at any section in the second span is 
M" x = M 2 + V 2 X—wx 2 , 

= —-£%wl 2 -\-^wlx—wx 2 . 


VALUES OF MOMENTS AND SHEARS 


153 


The moment is maximum when x=^l, since this is the value that makes 
the shear in the second span equal to zero. The value of the maximum posi¬ 
tive moment in this span is tz^qwI 2 . 

(e) The greatest bending moment that the beam is subjected to is the 
negative moment at the second (or fourth) support, and hence the maximum 
unit-stress in the beam is 


s 


I 28 


c 

7 * 


(/) The elastic curve equation for the first span is 


d 2 y 


11 


EI—~ = R ix -= —wlx -. 

dx 2 2 28 2 


Integrating we have, 


„jdy 11 x 2 wx 3 

El—= —wl -f-ci 

dx 28 2 6 

_ T 11 X 3 wx 4 

Ely — —wl - \-c%x ~f- c 2 - 

y 28 6 24 


Now, when z = 0, y=0; and when x=l, y = 0; thus c 2 =0, and Ci may easily be 
found to be —-^wl 3 . Therefore the elastic curve equation of the first span is 


__ 11 x 3 wx 4 1 

Ely = —wl - wl 3 x. 

y 28 6 24 42 


59. Values of Moments and Shears.—In Fig. 137 are given 
the values of the coefficients from which the negative moments 

0_0 

ofi Ho 

0_g_0 

oil tfi ito 


0 

10 

_1_ 

10 

0 


.6 

t^ 

5 t JL 

jL.fi 

Hio 

10 

1 10 

10 1 10 

10 I* 


0_ fa _ ~£a _ fs _0 

oS mm mm m 

0_^_£^3_ fa _^_0 

ofS iffi ifSt iffif iSi iifo 

Fig. 137.—Values of shear and moment coefficients for continuous beam of 
equal spans subjected to uniform load. 


over the supports and the vertical shears at either side of each 
support may be found for continuous beams resting on supports 
on the same level, having equal spans of l ft. each, and carrying a 
constant uniform load of w lb. per ft. on each span. 








154 


STATICALLY INDETERMINATE BEAMS 


The values directly above the supports are the moment coeffi¬ 
cients, that is, the numbers by which — wl 2 must be multiplied 
to obtain the bending moments over the supports. Similarly, 
the numbers on either side of the supports are the shear coefficients, 
that is, the numbers by which wl must be multiplied to obtain the 
magnitude of the vertical shear for the section; the vertical 
shear is negative for the section to the left of each support and 
positive for the section to the right of each support. The reaction 
at each support is the arithmetic sum of the vertical shears on the 
two sides of the support. The values given in Fig. 137 may be 
found by the methods discussed in Art. 58. 


PROBLEMS 

144. A continuous beam consists of four equal spans each 12 ft. long, and 
is subjected to a uniform load of 100 lb. per ft. over its entire length, (a) 
Find the reaction of the second support. ( b ) If the beam has a rectangular 
cross-section, the depth being twice the width, what is the depth if the max¬ 
imum fiber unit-stress in the beam is 100 lb. per sq. in.? 

Ans. (a) #2 = 1370 lb., (6) depth = 6.06 in. 

145. A standard 12-in. 40-lb. I-beam is used as a horizontal continuous 

beam over four spans. The lengths of the spans are 12, 16, 16 and 12 ft. 
respectively, from left to right. The loads are uniformly distributed over 
each span, the loads per foot of length on the four spans being 1600, 2400, 
2400, and 1600 lb. per ft., respectively. Find (a) the maximum fiber unit- 
stress in the beam. Ans. (a) s = 16,630 lb. per sq. in. 

146. A continuous beam has three spans of 10 ft. each. The first span 
is subjected to a uniformly distributed load of 3000 lb. per ft. The other 
spans are not loaded. Find the four reactions. 

Ans. #! = 13,000 lb. # 2 = 19,500 1b. # 3 =-3000 1b. # 4 = 5001b. 

147. What should be the depth of a pine continuous beam having four 
equal spans of 7 ft. in order to resist a uniformly distributed load of 300 lb. 
per ft. over each span? Use a working unit-stress of 1000 lb. per sq. in. and 
make the depth of the beam twice the width. 

148. A 10-in. 25-lb. I-beam is continuous over three spans, each having a 
length of 16 ft. The first and third spans are loaded with 800 lb. per ft. and 
the second span is not loaded. Find the maximum fiber unit-stress in the 
beam and draw, to scale, the moment and shear diagrams. 

149. A beam is continuous o\ er three spans of 20,10, and 15 ft. respectively. 
The first and third spans are each loaded by a uniformly distributed load of 
400 lb. per ft. and the second span is not loaded. Find the moment over each 
support and the shear at the right of the first support. 


THEOREM OF THREE MOMENTS 


155 


60. Advantages and Disadvantages of Continuous Beams.— 

The remarks in Art. 56 concerning fixed-ended beams apply 
also, in the main, to continuous beams; the negative moments 
over the supports reduce the moments (and hence stresses) near 
the centers of the spans, and hence continuous beams are stronger 
and stiller than simply supported beams of equal spans. On the 
other hand, the uneven settlement of supports may change the 
moments at the supports and throughout the beam from those 
found by the analysis given above; further, the stiff er the beams 
the greater is the change in the moments. Partly for this reason 
and partly because the loads on the various spans of continuous 
beams in structures may vary considerably from that assumed in 
the design, it is frequently assumed in practice that in uniformly 
loaded beams of four or more equal spans all the spans (except the 
end spans), are subjected to the same maximum bending moment, 
a common value being +%wl 2 ; the maximum moment in the end 
spans is somewhat greater (t+ivI 2 is frequently used). Compare 
these values with those given in Fig. 137 for beams having more 
than four spans and note that the moments do not vary greatly 
over the inner supports. Two-span and three-span beams are 
stressed higher than beams of four or more spans, and for such 
beams the maximum moment in each span is frequently assumed 
to be -Yowl 2 . However, the values used in any case may be based 
on the results of an analysis of the moments in continuous beams 
similar to that discussed in the preceding articles. 

61. Theorem of Three Moments for Concentrated Loads.— 
The loading on most continuous beams may be assumed to be 
uniformly distributed without introducing serious errors, but in 
some cases heavy concentrated loads act on the beams and the 
results given in Art. 59 are not applicable. Space does not permit 
the derivation, but the theorem of three moments that applies to a 
beam of constant cross-section with supports on the same level 
and subjected to concentrated loads is as follows: 

M ih+ 2 M2(11+12) +M3I3 

= —Pih 2 (k — k s ) — P 2 / 2 2 (2A; — 3k 2 +k 3 ). (118) 

Mi, M 2 and Ms are the bending moments at any three consecu¬ 
tive supports, h and I 2 being the lengths of the two consecutive 
spans at the ends of which the moments are Mi, M 2 and M 3 . 


156 


STATICALLY INDETERMINATE BEAMS 


P i is a concentrated load in the first of the two spans at the dis¬ 
tance kli from the first of the three supports, and P 2 a concen¬ 
trated load in the second of the two spans at the distance kh 
from the second of the three supports, where k\ and ^2 are pro¬ 
portional parts of the spans such as J, J, etc. 

PROBLEM 

150. A continuous beam has two spans, the left span 20 ft. long and the 
right span 18 ft. long. The 20-ft. span has a load of 1600 lb. 10 ft. from the 
left end, and the 18-ft. span has a load of 15,000 lb. 8 ft. from the right end. 
Eind the proper size of a steel I-beam to resist these loads using a working 
unit-stress of 16,000 lb. per sq. in. Ans. 12-in. 40-lb. I-beam. 


CHAPTER VIII 


DEFLECTION OF STATICALLY DETERMINATE BEAMS 

(Moment-area Method) 

Note. Arts. 44 to 46 should be studied before this chapter is 
read. 

62. Introduction. —In Chapters VI and VII the elastic curve 

(P v 

equation M=±EI-^~ was used (1) to find the deflection of 

beams and (2) to find the reactions (both forces and couples) of 
beams for which the number of unknown reactions was greater 
than the number of equilibrium equations, that is, of statically 
indeterminate beams. The method of treating the equation in 
each case was to express M in terms of the reactions and x, and then 
to integrate both sides of the equation, determining constants 
in the equations (whether constants of integration or reactions) 
from the physical conditions satisfied by the beam. This method 
was called the double-integration method. 

Another method of treating the equation leads to the use of 

moments of areas of the moment-diagram ^or of the ^ -diagram j 

for finding deflections of beams and reactions of statically indeter¬ 
minate beams, and hence the method is called the moment-area 
method. For various purposes and for certain types of beams, as 
for example beams having a variable moment of inertia, it possesses 
advantages over the double integration method although it should, 
perhaps, be considered supplementary to, rather than a substi¬ 
tute for, the double integration method. 

Two methods of using moments of areas of the moment- 

diagram ^or of the ^ -diagram^ will be discussed briefly; namely, 

the slope-deviation method and the conjugate beam method. 1 

1 See “ Deflections of Beams by the Conjugate Beam Method," by H. M. 
Westergaard, Journal of the Western Society of Engineers, vol. 26, Nov., 
1921. 


157 


158 DEFLECTION OF STATICALLY DETERMINATE BEAMS 


These two methods will be used in this chapter for determining 
the deflection of statically determinate beams, and the same 
two methods will be used in the following chapter for determining 
the reactions, as well as the deflections, of statically indeterminate 
beams. 

Slope Deviation Method 


63. Theorems of the Slope-deviation Method. —As explained 
in Art. 45, the equation of the elastic curve of a beam is 

M= —EI~^, in which M is positive when it produces tensile 

stress on the bottom of a horizontal beam, and y is positive when 
measured downwards. This equation may be written 


M_ 

El 


,(dy\ 

\dx) 


dx 


or 


K dx 

EI 


-"(I): 


(119) 


dy . 


in which is the slope of the elastic curve at a point for which 

the bending moment is M. If 0 denotes the angle made by the 
tangent line to the elastic curve with the axis of the undeflected 


beam, then ^ = tan 
dx 


But in a well designed beam 0 is small, 


and hence, without appreciable error, tan 0 may be replaced by 0. 
Therefore, 


dO 


EI 


dx and Ad 


/ 


dx, 


( 120 ) 


in which AO denotes the change or increment in the angle 0; 0 
is considered positive when measured in the clockwise direction 
of rotation. 

A graphical interpretation of this equation may be made as 
follows: Fig. 138(a) represents a beam subjected to a distributed 

M 

load; Fig. 138(6) represents the -^-diagram for the beam, any 

ordinate in which is the bending moment at the section where the 
ordinate is erected divided by EI; if the beam is homogeneous 
and has a constant cross-section, EI is a constant and hence the 
M 

•gj-diagram has the same form as the moment-diagram. In 
Fig. 138(c) PABQ represents the elastic curve of the beam. 



THEOREMS OF THE SLOPE-DEVIATION METHOD 159 


Now 


M 


indicated in Fig. 138(6), jdx is represented by an 
M f B M 

elementary area under the -^-diagram and J is repre- 

M . A 

sented by the total area under the -^-diagram between specified 

HI 

ordinates, such as ordinates erected at A' and B'. Further, dd 
represents, as shown in Fig. 138(c), the change in the slope of the 
elastic curve at two points a 
short distance, dx, apart, and 
m 0 A 


If :A 


TB 0 

K 1 

^ i 


(a) i A 

| 

1 

i 

i 


[j-^ — diagram 

1 S' 

i 

T 

l / 

1 

i : N 

r 1 


A'i 

A i ax 

(6) ! 8 ' 

1 

! Q 



hence I ‘ dd = Ad represents 

JQb 

the change in slopes of the 
elastic curve at any two points, 
such as A and B. Therefore, 
equation 120 leads to the fol¬ 
lowing theorem: 

Theorem I. When a straight 
beam is subjected to bending the 
difference in the slopes of the 
elastic curve at any two points 
is equal in magnitude to the 

area of the diagram between Fl «- 138.—Deflection of a beam; slope- 
B1 deviation method. 

the two points. 

Now let t, Fig. 138(c), denote the distance of any point A on 
the elastic curve, measured in a direction perpendicular to the 
original position of the beam, from a tangent drawn at any other 
point B on the elastic curve. The distance t will be called the 
tangential deviation of the point. Then, from Fig. 138(c), 


=f dt = f xde > .( 121 ) 


and from equation 120, 


Therefore, 


M , 

dd = ^jdx. 


CMx , 

=J w dx - 


( 122 ) 


Mx 7 . 


But, as is evident from Fig. 138(6), jdx is the moment of the 















































160 DEFLECTION OF STATICALLY DETERMINATE BEAMS 


elementary area, 


Mdx 

~WT' 


M 

of the -^-diagram about the ordinate 
JcjI 


through the point A whose tangential deviation is t, and hence 

/ Mx M 

-jjrjdx is the moment of that part of the -^-diagram between 


the two ordinates considered; the moment being taken about 
the ordinate through the point whose tangential deviation is 
desired. The following theorem therefore may be stated: 

Theorem II.—When a straight beam is subjected to bending the 
distance of any point A on the elastic curve, measured normal to the 
original position of the beam, from a tangent drawn to the elastic 
curve at any other point B, is equal in magnitude to the moment of 


M 

the area of the r—^-diagram between the two points about an ordinate 


through A. 


Applications 


64. Simple Beam. Concentrated Load at Mid-span. —The 

beam is shown in Fig. 139(a); the weight of the beam is assumed 

to be negligible. Fig. 139(6) 



shows the -^j-diagram; the 

dangerous section is at the mid- 
span; the maximum bending 
PI 

moment is (Art. 36), and 

the bending moment at a sec¬ 
tion the distance x from the 

p 

left support is ^ x . Since the 

beam is assumed to be homo¬ 
geneous and to have a constant 

cross-section, the ^-diagram 

has the same form as the mo- 
In Fig. 139(c) 
ACB represents the elastic 
curve of the beam. 


D 

Fig. 139. —Deflection of simple beam, ment-diagram. 
load at center; slope-deviation method. 


Let it be required to find the maximum deflection A, which 
occurs at the center of the span. The tangential deviation of 
the point A from a tangent at C is t A (Fig. 139c) which is equal in 

















CANTILEVER BEAM 


161 


magnitude to A, and which may be found from Theorem II of 
Art. 63. Thus, 

A = t A = Moment of area A'C'H about A' 

= Area A'C'H times distance to centroid, G, of area 
A'C'H from A' 

= 1 PI J 21 = PI* 

2 4#/2 3 2 48 EP 

Next let it be required to find the deflection y of a point E 
(Fig. 139c) at a distance x from the left support. If a tangent 
AD is drawn to the elastic curve at A, is the tangential devia¬ 
tion of B and t\ is the tangential deviation of the point E. From 
the geometry of the figure we have, 

y+ti x . x . 

__ = r or y=h T -t u 

and since the slopes of the elastic curve are small we have, 6 
Therefore, 

y = dx—t\. 

Now 9 is equal to the change of the slopes at A and C and hence, 
according to Theorem I, 

0 = area A'C'H 
_ 1 PI l _ 1 PI 2 
2 4 El 2 16 El ’ 

and, according to Theorem II, 

ti= moment of area A'MN about MN 

1 Px X x_ Px 3 
~2EI 2 3 — 12#7‘ 

Therefore 

_Pl 2 x Px 3 
V ~16EI 12 El’ 
or 

48EIy=Px(Zl 2 -4:X 2 ), 

which is the elastic curve equation for the left half of the beam. 

65. Cantilever Beam. Concentrated Load at End.—In 

Fig. 140(a) AB represents the elastic curve of the beam, and in 

M 

Fig. 140(6), A'B'H represents the -gj-diagram. The beam is 
assumed to have a constant cross-section and the weight of the 


h 

V 








162 DEFLECTION OF STATICALLY DETERMINATE BEAMS 


beam is neglected. The maximum deflection of the beam may 
be found as follows. 

The tangential deviation, t Aj of A from a tangent drawn at B 
is^equal to the maximum deflection A of the beam. Thus from 
Theorem II we have, 

A = t A = moment of area A'B'H about A' 

— Area A'B'H times distance to centroid. 

1PP 2 1 PE 

2 El'3 l 3 El ’ 


The deflection, y, of a point D at the distance x from the free 
end may be found as follows: The tangential deviation, t Df of D 
from a tangent at B is equal to y; thus, using Theorem II we have, 

2 / = ^=moment of area B'D'EH about D'E. 


Let the area be divided into two triangular areas as indicated 
by the dotted line. Then 


Hence, 




«-»>■ | « 

3 + 2 ei :v ; ' 


6EIy = P(x 3 -3l 2 x+2l 3 ) 


W lb. per ft. 



Fig. 140.—Deflection of cantilever Fig. 141.—Deflection of simple 
beam, load at end; slope-devia- beam; uniform load; slope- 

tion method. deviation method. 


which is the elastic curve equation for the beam. 

66. Simple Beam; Load Distributed Uniformly.—The beam 
shown in Fig. 141(a) has a constant cross-section and the weight 
of the beam is negligible. The moment diagram A'HB' (Fig. 






























CANTILEVER BEAM 


163 


1416) is a parabola, the ordinate M to any point being expressed 
by the equation M = \wlx—\wx 2 , and the maximum ordinate 
being \wl 2 (see Prob. 88). The elastic curve is represented by the 
line ACB in Fig. 141(c). 

The maximum deflection of the beam may be found as follows: 
The tangential deviation t A of the point A (Fig. 141c) from a 
tangent at C is equal in magnitude to the maximum deflection, A, 
of the beam. Hence according to Theorem II we have 


A = t A = -^j (moment of area A'C'H about A '). 

Mil 

Now, 

Area A'C'H = iA'C'XC'H (see Art. 159, Appendix II), 
and the distance of the centroid of area A'C'H from A' is f AC'. 
Therefore, 

A _, _ 1 /2 1 12 l 5 l \ 

A A EI\Z% Wl 2 8 2/ 

_ 5 wE_ 5 WP 
384 El 384 EE 


67. Cantilever Beam; Load Distributed Uniformly.—The elas¬ 
tic curve of the beam is repre¬ 
sented by the line AB (Fig. 
142a), and the moment-dia¬ 
gram for the beam is shown in 
Fig. 142(6). The maximum 
moment, at the wall, is — \wl 2 
and the moment at a distance 



x from the free end is — 

Now the tangential deviation 

„ T*-. n i. , of the point A from a tangent 

Fig. 142. — Deflection of cantilever n ■ . . 

beam, uniform load; slope-deviation drawn at B is equal m magm- 
method. tude to the maximum deflec¬ 

tion, A. Hence, by integrating 

equation (122), we have 
A 


-AX 


Mxdx 


wl 4 

8 El 


~ 2 ~.xdx 

1 WP 

8 EF 















164 DEFLECTION OF STATICALLY DETERMINATE BEAMS 


or, from Theorem II, the value of A is found without integrating: 

A = t A = r— (moment of area A'B'H about A') 
til 


__1_/1 wl 2 3 
~EI\ 3 2 

_ 1 wl 4 

~8 ~Tr 


(See Art. 159, Appendix II) 


Conjugate-beam Method 

68. Conjugate Beam Defined. Equations Stated.—By the 

conjugate-beam method the deflection at any section of a beam 
(here called the “ given ” beam) is found by calculating the 
bending moment at the corresponding section of another beam 
(here called the conjugate beam), the conjugate beam being sub¬ 
jected to a distributed load such that the intensity of the load at 

M 

any section is proportional to the ordinate of the ^-diagram at 

til 

that section. That is, the conjugate beam may be assumed to be 
loaded with sand the depth of which over any section of the beam 

M 

is, according to a certain scale, the ordinate of the ^-diagram 

til 

at that section. To describe this loading the conjugate beam will 

M 

be said to be loaded with the -^-diagram. 

til 

By definition the conjugate beam is one that (a) has a length, 
V, which is equal to the length, l, of the given beam, (6) is in 
equilibrium, and (c) is so loaded that the bending moment M r 
at any section is equal to the deflection, y, at the corresponding 
section in the “ given ” beam. The defining equations then are: 
From (a) 

V = l; 

from ( b ), since the loads and reactions constitute a parallel system 
of vectors, 2 there are two equations of equilibrium; namely, 

2F' = 0 and 2Af' = 0 

and from (c) 

M' x = y x . 

2 It will be found that the loads and reactions are not forces; that is, they 
are not expressed in pounds, tons, etc., but for convenience they may be called 
elastic forces, since they involve the elastic properties of the beam. 


CONJUGATE BEAM DEFINED 


165 


The last equation requires, as is shown below, (1) that the vertical 
shear, V', for any section of the conjugate beam shall be equal to 
the slope at the corresponding point of the elastic curve of the 
“ given ” beam; that is, 



(123) 


and (2) that the intensity, w', of the distributed load on the con- 

M 

jugate beam at any section shall be equal to the for the 

til 

corresponding section in the given beam; in other words, the 

M 

conjugate beam is loaded with the ^^diagram. Thus, 

hii 



(124) 


Proof of Equations (123) and (124).—Let a straight beam be 
subjected to a distributed load only; the intensity of the load at 
the distance x from the left support is w x , and the vertical shear 
and bending moment for this section are V x and M x . In the 
following equations the sign of the bending moment is determined 
as stated in Art. 34; the vertical shear V x is positive when 
directed upwards; the deflection y is positive in the downward 
direction, and x is positive to the right, and hence the elastic 



when measured in the clockwise direction of rotation. 

Now the difference, dV, in the vertical shears for two sections 
the distance dx apart is 


dV = wdx or w— j— 
dx 


(125) 


But from Art. 37, 



dx ’ 


(126) 


and hence, 


dV __ d 2 M _ 
dx dx 2 W ' 


( 127 ) 


Now the elastic curve equation of the beam is (see Art. 45), 







166 DEFLECTION OF STATICALLY DETERMINATE BEAMS 


Thus a comparison of equations (127) and (128) shows that the 
deflection y of the “ given ” beam would be equal to the bending 
moment M' at the corresponding section of another beam (the 

M 

conjugate beam) which has a distributed load equal to that 

M 

is, the conjugate beam is loaded with the -gj-diagram. 

Further, since the slope 0 of the elastic curve at any point is 
6 = ~, it follows that, if the conjugate beam is loaded so that 

(XX 

M' = y, then from equation (126) 



That is, the vertical shear for any section in the conjugate beam 
must be equal to the slope of the elastic curve at the corresponding 
section in the given beam. 

It will be noted that, in the conjugate beam method, the 
actual elastic beam is replaced by a rigid beam with the elastic 
properties of the actual beam introduced in the loads and reac¬ 
tions of the rigid beam. For this reason the method is some¬ 
times referred to as the method of elastic weights. 

Summarizing: The equations, then, that the conjugate beam 
for any given beam must satisfy are 

l' = l, SF' = 0, 2M' = 0, M' = y 

r=e, »'=^. 

The applications of these equations in finding the deflection of 
various types of beam is given below. 


Applications 


69. Simple Beam; Concentrated Load at Mid-span.—The 

beam is bent as shown in Fig. 148(a). The bending moment dia¬ 
gram (M-diagram) is shown in Fig. 143(6) and since the beam has a 


constant cross-section the 


M_ 

El 


diagram will have the same form 


as the Af-diagram. The conjugate beam is shown in Fig. 143(c); 
the bending moment at the mid-span section of the conjugate 
beam is equal to the deflection of the mid-span section (maximum 


SIMPLE BEAM, CONCENTRATED LOAD 


167 


deflection) of the given beam, provided that the conjugate beam 
is made to satisfy equations of Art. 68. This may be done by 
selecting the beam as shown in Fig. 143(c) in which l' = l; the dis- 
M 

tributed load is the ^-diagram; since M' = y , and since y = 0 at 
hi 

the ends of the “ given ” beam, the moment M' at the ends of the 
conjugate beam must also equal 
zero, and hence the ends of the 
conjugate beam are subjected to 
zero moments, that is, the ends 
are free to turn as shown in 
Fig. 143(c); further, since V' = d 
and since 0 is not zero at the ends 
of the given beam, there must 
be vertical shears at the ends of 
the conjugate beam, and these 
could be produced by reactions 
R'i and R '2 of supports at the 

ends. Thus, if the “ given” beam . - ^ 

is a simple beam the conjugate ^ 143 . _ Deflection of C simple 
beam also is a simple beam. beanl) load at center . conjugate _ 

Now the maximum deflection beam method. 

A of the given beam occurs at 

the mid-span section and hence it is equal to the moment, M' c , 
at the center of the conjugate beam, which is the moment of the 
couple shown in Fig. 143(c). Thus, 

, _J [_Pl 2 2 l_ _1 JPI* 

A ~ M c 16 El 3 2 48 EF 

Further, since the conjugate beam is in equilibrium the values 
of R'i and R '2 may be found from the equations 2F' = 0 and 
2AP = 0. Thus, from 2Af' = 0, or from the conditions of sym¬ 
metry, we obtain, R'i = R r 2 . And from 2F' = 0 we obtain the 
values of R'i and R' 2 , which are equal to the vertical shears at 
the ends of the conjugate beam and hence equal in magnitude to 
the slopes, d A and 0 B , at the ends of the “ given ” beam. Thus, 

* F "= 2R, '-\m- l = Q ' 



1_PP 
16 EV 


whence, 






























168 DEFLECTION OF STATICALLY DETERMINATE BEAMS 


as is evident from inspection of Fig. 143(c). Therefore, 

_J _Pl 2 
° A ~m Et 

Deflection at Any Point. —The deflection y of the u given 
beam at a distance x from the left support is equal to the bending 
‘moment M' x of the conjugate beam which, as indicated in Fig. 
144, is 

D , lPx 2 1 
y = M' x = R \X - 2 -y • 

_ Pl 2 x }_Px^ = P /Px_x^\ 

~mi 12 El ~4:EI\ 4 3/* 



load at end; conjugate-beam method. 


This is the equation of the elastic curve of the left half of the 
given beam. 

70. Cantilever Beam; Concentrated Load at End.—The 

given beam is assumed to have a constant cross-section; it bends 
as shown in Fig. 145(a). The conjugate beam is shown in Fig. 
145(6); the distributed load on the conjugate beam is the 
M 

-pry-diagram and is an upward load since M is negative. The 
hi 

end conditions of the conjugate beam are found by making the 
beam satisfy the fundamental equations of Art. 68. Thus, since 
M' = y and V' = 0, and, further, since y and 0 are not zero at the 
free end of the “ given ” beam, then the moment M'o and the 
shear V'o at the left end of the conjugate beam are not zero. 
M'o and V'o may be produced by fixing the beam at the left end. 
On the other hand, since the 6 and y at the right end of the “ given ” 
beam are zero, the M' and V' at the right end of the conjugate 
beam must be zero, and this condition will exist if the right end is 

























SIMPLE BEAM, UNIFORM LOAD 


169 


free. Therefore, the conjugate beam corresponding to a canti¬ 
lever beam is another cantilever beam with the end conditions 
interchanged. 

Now the maximum deflection, A, of the “ given ” beam is 
equal to the moment M'o at the left end of the conjugate beam. 
Thus, 


A = M' 


1 Pl 2 _ 2 

2 EI S 1 


1 PE 
3 EI‘ 


The slope, 0, at the free end of the given beam may be found, 
if desired, by calculating the shear V' 0 at the left end of the con¬ 
jugate beam; the value found is 


d = V' 0 


1 PE 

2 EE 


The elastic curve equation may be found in a manner similar 
to that used in the preceding article. 



Fig. 146.—Deflection of simple 
beam, uniform load; conju¬ 
gate-beam method. 


Fig. 147.—Deflection of cantilever 
beam, uniform load; conjugate- 
beam method. 


71. Simple Beam; Load Distributed Uniformly.—The beam 
is assumed to have a constant cross-section; it deflects as shown 
in Fig. 14.6(a). The corresponding conjugate beam is shown in 

M 

Fig. 146(6), the distributed load being the -^j-diagram, which is 
a parabolic area, the maximum ordinate of the diagram being 
~ (see Prob. 88). 

The maximum deflection A is equal to the moment at the 
center of the conjugate beam, which is the moment of the couple 
shown in Fig. 146(6). Thus, 






































170 DEFLECTION OF STATICALLY DETERMINATE BEAMS 


A , jr _2 1WI l 51 
A Mc 38 El'2'8 2 

= _5 wP = _5__WP 
384 El 384 El * 


72. Cantilever Beam; Load Distributed Uniformly. —The 

beam has a constant cross-section; it deflects as shown in Fig. 
147(a). The corresponding conjugate beam is shown in Fig. 
147(6). The maximum deflection A is equal to the moment, 
M' 0 , at the left end of the conjugate beam. Thus, 

1 wl^ 3,1 wP _ 1 Wl 3 
A ~ M ° Z 2EI & 8 El 8 EI~ 


73. Simple Beam; Concentrated Load at Any Point.—Let 

the load act at a distance a from the left end and 6 from the right 

M 

end; the deflected beam is shown in Fig. 148(a). The -^-diagram 

is shown in Fig. 148(6) as a distributed load acting on the con¬ 
jugate beam; the “ given ” beam is assumed to have a constant 
cross-section. 

The total load is ^ • Z and its action line is at a distance of 

z hilt 


§(Z+a) from the left end. From the equations of equilibrium, 
XF' = 0 and 2Af' = 0, we find the values of R'i and R '2 (and hence 
of 6 a and 6 b ) to be, 


p r 1 l~\~b Pcib 

Ul= z~rm 


and 


R' 2 


1 Z+a Pab 


The deflection y at the distance x from the left end of the given 
beam is equal to the bending moment, M' x , in the conjugate 
beam. Thus, 

y=M x =R,x-^- wl x 

_ 1 l+b Pab 1 Pbx 3 
3 l 2 EI X 6 IEI 
Phr 

.... (129) 

which is the elastic curve equation of that portion of the beam 
to the left of the load P. 

The maximum deflection may be found as follows: The max- 
mum moment in the conjugate beam is equal to the maximum 


CROSS-SECTION NOT CONSTANT 


171 


deflection in the given beam; now the section of maximum moment 
in the conjugate beam is the section for which the vertical shear in 
the conjugate beam is zero. Thus, the value of x' in the following 
equation locates the point of maximum deflection in the given 
beam. 


V' 


jj, Pbx 2 . 

Rl ~mr Q 


whence, 


_ 1Z+6 Pab Pbx 2 _ _ 

~z~r2Ei~wn~"' 

x = Vfr(J+b) .(130) 


If this value of x is substituted in equation (129), the resulting 
value of y is the maximum deflection A, (see Art. 50 for the value 
of A). 

PROBLEM 


151. Find, from equations (129) and (130), the maximum deflection of a 
simple beam subjected to a concentrated load at the center of the span, and 
compare the result with that given in Art. 47. 



Fig. 148.—Deflection of simple beam, Fig. 149.—Deflection of beam with 
load at any point; conjugate-beam variable section; conjugate-beam 
method. method. 


74. Simple Beam; Cross-section not Constant.— Let a con¬ 
centrated load act at the center of a simple beam, and let the 
moment of inertia for any section in the center half of the beam 
be I and for any section in the outer quarters \l as indicated 

































































172 DEFLECTION OF STATICALLY DETERMINATE BEAMS 


in Fig. 149(a). The moment diagram is shown in Fig. 149(6), 

and the ^-diagram is shown in Fig. 149(c) acting as a load on 
tLl 

M 

the conjugate beam. The -^-diagram may be divided into four 

hi 

triangles, I, II, III and IV, as shown in Fig. 149(c). From the 
equilibrium equation 2F' = 0, we have 

5 PI 2 

#'l =Ql+Q2 = gHrp 

and the maximum deflection A, which is equal to the bending 
moment M'c at the center of the conjugate beam is, 

A=M’c— R'i-—Qi^l—Q2^l 

_ _l_ _ 1 \PP 

\128 96 192/ El 

3 PP 
~ 128 El' 

PROBLEM 

152. A simple beam is subjected to a concentrated load P at the center 
of the span; the moment of inertia of the left half of the beam is / and of the 
right half is \I. Derive the expression for the maximum deflection of the 
beam. 

Note. —The problems in Chapter VI may also be used in connection with 
this chapter. 


CHAPTER IX 


STATICALLY INDETERMINATE BEAMS 

(Moment-area Method) 

75. Introduction. —As explained in Art. 51, a statically inde¬ 
terminate beam is one for which the number of reactions is greater 
than the number of equations of equilibrium; in order to deter¬ 
mine the reactions for such a beam the equation of the elastic 
curve of the beam is needed, -in addition to the equations of equi¬ 
librium. The use, by the double integration method, of the gen¬ 
eral equation of the elastic curve for this purpose, is discussed in 
Chapter VII. In this chapter the moment-area method will be 
used to furnish sufficient equations, in addition to the equations 
of equilibrium, to determine the reactions of statically indeter¬ 
minate beams. Further, two methods of using moment-areas 
will be employed; namely, the slope-deviation method and the con¬ 
jugate-beam method; the equations and theorems used in the 
slope-deviation method are stated in Art. 63, and the equations 
used in the conjugate-beam method are given in Art. 68. 

Slope-deviation Method 

76. Beam Fixed at Both Ends; Load Concentrated at Mid¬ 
span. —The beam is shown in its deflected form in Fig. 150(a). 
The beam may be considered to have been a simple beam to which 
negative end-moments have been applied which cause zero slopes 
at the ends as indicated in Fig. 150(5). The moment-diagram 
may be found by superimposing the positive moment-diagram 
for a simple beam subjected to a concentrated load at the mid¬ 
span, the maximum ordinate to which is at the mid-span and 

PI 

equal to -j-, and a negative moment-diagram consisting of a 

rectangle, the constant ordinate to which is the unknown end- 
moment Mo; this superimposed diagram is shown in Fig. 150(c), 

173 


174 


STATICALLY INDETERMINATE BEAMS 


the end-moments being known to be equal from the conditions 
of symmetry (or of equilibrium). Since the beam has a constant 

M 

cross-section, Fig. 150(c) also represents the -^-diagram. Since 
the beam is fixed at both ends, we have 

Change of slope from A to B = zero, 
and hence, from Theorem I of Art. 63, we have 


Total area of ^-diagram = 0, 

LjI 

that is, 

"■+«•)-o. 

and hence, 

M 0 =-\Pl. 



Fig. 150.—Moments at ends of fixed Fig. 151.—Moments at ends of fixed 
beam; slope-deviation method. beam; slope-deviation method. 


According to Theorem II of Art. 63 the maximum deflection, 
A, which occurs at C (Fig. 150a), is 
M 

A = moment of -^-diagram from A' to C' about C 
Jhl 

1 (moment of area A'DC' about C'D-\- moment of 
~ El area A'GEC' about C'E) 

= J J_\Pl l_ 1 1 , PI l_ l\ 

EI\ 2 4 *2*3 2' 8 *2*4/ 

PP 


192 EP 





































































BEAM FIXED AT ONE END 


175 


77. Beam Fixed at Both Ends; Load Uniformly Distributed. 

The deflected beam is shown in Fig. 151(a). The ^-diagram 

El 

is shown in Fig. 151(6); it is considered to be composed of two 
parts, as in the preceding article, consisting of the positive area 
A'DB' and the negative area A'GHB'. 

Since the change in slope from A to B (Fig. 151a) is equal to 
zero we have, from Theorem I of Art. 63, 


that is, 


Hence, 


Total area of 


El 


-diagram = 0, 


2 MoZ_ 

3 8EI t+ El~ 


Mq — — 


wl 2 

12 * 


And from Theorem II of Art. 63, the maximum deflection, A, is 
A = moment of area A'DC' about CD —moment of area 
A'GEC' about C'E 

_2 wl 2 l 3 l / wl 2 \ l 
3 SEI 2 w' 2\12EI / 4 

wl 4 

“384 El' 


78. Beam Fixed at One End, Supported at Other End; Load 
Concentrated at Mid-span. —The deflected beam is shown in 
M 

Fig. 152(a), and the -^-diagram considered to be made up of 
El 


two parts, as discussed m Art. 76, 
is shown in Fig. 152(6). Thus 
the moment-diagram consists of 
a triangle with the maximum 
ordinate at the center of the span, 
the same as that for a simple 
beam with a concentrated load P 
at the mid-span, and a negative 
triangular diagram the ordinates 
to which vary from Mo at the 
wall to zero at the left end. Fi 



beam; slope-deviation method. 

er, since the cross-section of 



















176 


STATICALLY INDETERMINATE BEAMS 


the beam is constant, the ^-diagram has the same form as the 
El 


moment-diagram. 

Since the tangential deviation, t A , of the point A from a tan¬ 
gent drawn at B is equal to zero, we have, from Theorem II of 
Art. 63, 


t A = moment of 


M_ 

Ei 


■diagram about A' = 0 


of area A'DB' about A'+moment of area 
A'GB' about A') =0 

1 PI 11 Mol 2 
24EI ' l 2^2 EI Z 

M 0 =-&Pl 


= (moment 


Hence 



Fig. 153.—Moment at fixed end of 
beam; slope-deviation method. 


79. Beam Fixed at One End, 
Supported at Other End; Load 
Uniformly Distributed. —The de¬ 
flected beam is shown in Fig. 
M 

153(a) and the -^j-diagram is 

shown in Fig. 153(6). The tan¬ 
gential deviation of the point A 
from a tangent at B is equal to 
zero. Hence, from Theorem II 
of Art. 63 we have 


M 

t A — moment of ^-diagram about A' = 0 
El 


= moment of area A'DB' about A'+moment of area 
A'GB' about A' = 0 


Hence, 


2 wl 2 l_,lMo 2. 

3 8F/‘^2 + 2 EI 1 ' 3 1 


Mq— — 


wl 2 

T* 


80. Continuous Beam; Theorem of Three Moments. —Let 

Fig. 154(a) represent a continuous beam subjected to uniformly 
distributed loads; the beam has a constant cross-section and rests 
on supports that are on the same level. 


















CONTINUOUS BEAM 


177 


The elastic curve of two adjacent spans of the beam is shown 
in Fig. 154(6). If, over the supports, hinges are introduced, as 
indicated in Fig. 154(c) and then external end-moments, Mi, M 2 
and Ms, are applied as shown, the beam will act in all ways as the 
original beam acts, provided that the values of Mi, M 2 and M 3 
are the values of the bending moments at the supports of the orig¬ 
inal beam. The moment diagram formed by superimposing the 
positive and negative moments, caused by the loading shown in 
Fig. 154(c) is represented in Fig. 154(d). 



FiG. 154.—Moments at supports of continuous beam; slope-deviation method. 


Let a tangent be drawn to the elastic curve at the support B 
(Fig. 1546); since the elastic curve is continuous over the sup¬ 
port, the tangent to the elastic curve at B is common to the two 
curves of the adjacent spans h and I 2 . 

Now, from Theorem II of Art. 63, the tangential deviation h 
of the point A from a tangent at B is 

h = (moment, about A', of area A moment 

£j L 


about A', of area A'GHB') 




















































































178 


STATICALLY INDETERMINATE BEAMS 


-h(5T , ''3+5"4+I"- , ''t) 

1 / will* MJi 2 M 2 h 2 \ 

EI\ 24 + 6 3 /’ 

Similarly, taking moments, about C' } of areas of the moment 
diagram for the next span, we have 

1 / 2 w 2 h 2 7 , 1 M j l 2 . 1 u 7 2 l 2 \ 

t2= m\3~T' h ' 2 + 2 Msh ' 3 + 2 M24h T) 


1 / W 2 I 2 4 , M 2 I 2 2 , Msl 2 2 \ 


Now, from similar triangles, ^-= 


— 
h ’ 


whence, 

wiZi 3 ikfiZi M2I1 _ /W21 2 3 , -M3Z2 - M 2 h\ 

~24~ +_ 6 _i " 3 l 24 + 6 + 3 / 




or, 

MlZl ^(Zl+k) ,M3?2_ WlZl 3 ^2 3 
6 ' 3 ^ 6 24 24 ‘ 


Therefore, 

Mili-\-2M 2 (li-\-l2) — 


wih 3 


W 2 I 2 3 

~4T' 


This equation expresses the theorem of three moments for a 
continuous beam subjected to uniform loads, the supports being 
on the same level. For use of the theorem see Art. 58. 


Conjugate-beam Method 

81. Beam Fixed at Both Ends; Load Distributed Uniformly. 

—The deflected beam is shown in Fig. 155(a). In Fig. 155(6) 
the beam is shown acted on by an equivalent force system which 
converts the beam into a simple beam subjected to end-moments 
Mo. The moment-diagram for the beam in Fig. 155(6), and hence 
also for the original beam, is shown in Fig. 155(c); the positive 
moment diagram of the vertical forces is a parabola the same as 
that for a simple beam, and the negative moment diagram of the 
end-moments is a rectangle. Further, since the beam has a con- 














CONJUGATE BEAM METHOD 


179 


M 

stant cross-section, the -^-diagram has the same form as the 
moment-diagram. 

Now, as discussed in Art. 70, the conjugate beam for a fixed- 
ended “ given ” beam is a 

W lb. per ft. 



fnnnri 


W lb. per ft. 


amrm-rm 






(f>) 


beam with free ends, that is, 

M 

the -^j-diagram, acting alone 

as a distributed load on the 
conjugate beam, must hold the 
conjugate beam in equilibrium 
as indicated in Fig. 155(d), 

M 

and hence the -^-load diagram 
HI 

must be so adjusted that the 
sum of all the loads is equal 
to zero (2F' = 0, Art. 68); that 
M 

is, in Fig. 155(c), the -^-load 

of the parabolic diagram must 

be equal and opposite to the 

M , . „ . , .. Fig. 155. —Moments at ends of fixed 

w loa,d of the rectangular dia- beam . conjugate . beam method . 



, ^rfim 

Trrm^ 

if] 



A 

l 

c ' (c) B 


2 

■p. Conjugate 

DM&, 

A [ 

M 

WW. 

b % 

id) \ 


gram. Hence 


whence, 


2 wl 3 Mo 

3 SEI * 1 El’ 

Mo= wl 2 . 


Maximum Deflection .—The maximum deflection A of the 
given beam is equal to the bending moment M' c at the center of 
the conjugate beam. In finding M' c the load to the left of the 
center may be considered to consist of two parts: the rectangle 
ACQN (Fig. 155d) acting as an upward load, and the parabolic 
area NTQ acting as a downward load; these two areas are equal 
in magnitude since the beam is in equilibrium. Taking moments 
about C' we have, 

_ 1 W t 2 l l 2 1 wl 2 l 3 l 
A ~ Mc 12 El'2 4 3 8 El 2 82 


wl 4 __ Wl 3 
384 El 384A7* 


































180 


STATICALLY INDETERMINATE BEAMS 


82. Beam Fixed at Both Ends; Load Concentrated at Mid¬ 
span. —The deflected beam is shown in Fig. 156(a). A force 
system that would produce the same deflections as occur in the 
“ given ” beam is shown in Fig. 156(6), and the superimposed 
moment-diagram is shown in Fig. 156(c). The conjugate beam 
(Fig. 156d) for a fixed-ended “ given ” beam has free ends (see 

M 

Art. 70), and hence the -^-diagram must hold the conjugate 

M 

beam in equilibrium; that is, the total -^-load must equal zero 

El 

(XF' = 0, Art. 68). Hence, 


whence, 


l+~- ~ — -1 = 0 
El +2 4 El 1 


Mq — — 4PI. 




Fig. 156.—Moments ac ends of fixed Fig. 157.—Moment at fixed end of 
beam; conjugate-beam method. beam; conjugate-beam method. 


The maximum deflection A is equal to the bending moment 
M' c at the center of the conjugate beam. Thus, 

A = M' c = (moment ACQN —moment NTQ) 

= 1 Pl_l_ l___ll Pl_ l 1 1 
8 El 2 4 2 4 El 2 3 2 
J _PE 
192 El • 
































































CONJUGATE BEAM METHOD 


181 


83. Beam Fixed at One End; Supported at Other End; 
Load Uniformly Distributed.— The deflected beam is shown in 
Fig. 157(a). The superimposed moment-diagram is shown in 
Fig. 157(6). 

Since the deflection and slope at the right end of the “ given ” 
beam are zero, the shear and moment at the right end of the 
conjugate beam must be zero, and hence the end is free (see Art. 
70); whereas, the left end of the conjugate beam has zero moment 
(M' = y = 0) but not zero shear ( V' = 6 ), and hence is supported 

at the left end, and is subjected to the distributed ^-load as 

El 

indicated in Fig. 157(c). 

The value of Mo may be found by applying one of the equa¬ 
tions of equilibrium (2AF = 0). Thus if moments are taken about 
the left end of the conjugate beam the moment of the triangular 
area A'B'D plus the moment of the parabolic area A'ED must 
equal zero. Hence, 

1 Moj 2 2 IwE l__ 

2FJ 3 H ~3'8FJ 2 ’ 

whence, 

M 0 =~^wl 2 . 

Now R'i (and hence 6) may be found, if desired, from the other 
equilibrium equation (SF' = 0). 

84. Continuous Beam; Theorem of Three Moments. —Fig. 

158(a) shows two spans of a continuous beam subjected to dis¬ 
tributed loads; the supports are on the same level, and the beam 
has a constant cross-section. If, over the supports, hinges are 
introduced and then end-momenfcs, M i, M 2 and M 3, are applied, 
as in Fig. 158(c), the beam will act in all ways as the original beam 
acts, provided that the value of Mi, M 2 , and M 3 are the values 
of the bending moments at the supports of the original beam. 

The superimposed moment diagram for the beam in Fig. 
158(c) (and hence also of the original beam) is shown in Fig. 
M 

158(d), and the ^-diagram will have the same form as that of the 
El 

M 

moment-diagram. The ^-diagram is the distributed load on 

the conjugate beam (Fig. 158c); now the conjugate beam has 
hinges and no supports at the points corresponding to the sections 


182 


STATICALLY INDETERMINATE BEAMS 


over the support of the original beam. This may be explained 
as follows: The deflection at the supports of the “ given ” beam, 
and therefore the moment at the supports in the conjugate beam, 
are zero; the slope of the given beam at a support may be different 
from zero (as indicated in Fig. 1586) but must have the same 



Fig. 158. —Moments at supports of continuous beam; conjugate-beam 
method. 


value immediately to the left and to the right of the point since 
the beam is continuous over the support, and therefore the shear 
in the conjugate beam may have a value other than zero but 
must have the same value immediately to the left and the right 
of the point. An unsupported hinge is the simplest arrangement 
by which this condition can be established. 

By expressing the fact that the ^-load must be so adjusted 



















































































































































CONJUGATE BEAM METHOD 


183 


that the reaction of the support (B say, Fig. 158e) is zero, the 
theorem of three moments is found as follows: Let it be assumed 
M 

first that all the -^-loads are acting downward, the reaction at 

jbi 

B then would be such that its moment about A would be equal 

M 

to the moment, about A, of the 777 -loads on span h, and its moment 

M 

about C would equal the moment about C of the -^j-loads on h. 

Therefore, writing the expression for the reaction and equating it 
to zero, we have 

1/2 wih 2 7 h . 1 ,, 7 2 7 , 1 ,, 7 1 7 \ 
ii\3* 8 ' ll ' 2 + 2 M21 '3 1+ 2 ¥l l '3 V 

I/2W2I2 2 7 ^2 1 11/ 7 2 7 . 1 ,, 7 1 7 \ n 

+ fe (3 ~8 2 ' 2 + 2 J ^ 2 2 '3 2 + 2^ 3 2 3 2 ) _0, 

whence 

Mih +2M 2 (h+h) +Mzl 2 = - iwih 3 - iw 2 l2 3 , 
which expresses the theorem of three moments. For use of the 
theorem see Art. 58. 

Note.—T he problems in Chapter VII may also be used in connection 
with this chapter. 


CHAPTER X 


COMBINED AXIAL AND BENDING LOADS. ECCENTRIC 

LOADS 

85. Introduction.—In the preceding chapters stresses caused 
by axial, torsional, and bending loads were found when the loads 
acted singly. Members of many structures and machines, how¬ 
ever, are subjected to loads of two or more of these types (or the 
actual loads may conveniently be resolved into loads of two or 
more of these types), and the unit-stress developed at any point in 
the member may frequently be found by assuming that the loads 
act independently and hence each load is assumed to produce the 
same unit-stress that it would produce if it were the only load acting 
on the member; these unit-stresses may then be combined, if the 
stresses are within the proportional limit, to obtain the actual unit- 
stress. This principle, called the principle of superposition, will be 
used in this chapter in determining the normal (tensile or com¬ 
pressive) unit-stress at a point in a member when the member is 
subjected to combined axial (tensile or compressive) and bending 
loads, and also in determining the shearing unit-stress when the 
member is subjected simultaneously to a central shearing load 
and a torsional load. 

86 . A Beam Subjected to an Axial End Load.—In Fig. 159(a) 
is represented a simple beam of length l feet subjected to a uni¬ 
formly distributed transverse load of w pounds per foot and a com¬ 
pressive axial load of Q pounds at its ends. Let it be required to 
find the maximum normal (tensile or compressive) unit-stress on a 
normal cross-section of the beam. The beam will be assumed to 
be short so that the deflection of the beam may be neglected with¬ 
out introducing serious error in the results, and hence the load Q 
may be assumed to be an axial load for each cross-section of the 
beam. Further, the weight of the beam is assumed to be negli¬ 
gible in comparison with the other loads on the beam. 

184 


BEAM SUBJECTED TO AN AXIAL END LOAD 


185 


The maximum axial unit-stress will occur at the mid-section 
since the unit-stress due to the transverse bending load is maxi¬ 
mum at the mid-section and the direct stress due to Q is the same 
for all sections. Now as indicated by the free body diagram of 
the left half of the beam in Fig. 159(6), if Q were the only force 
acting, the stress would be uniformly distributed on the cross- 
sectional area a, the unit-stress at any point in the area being si. 
The total stress would be asi and would be equal and opposite to 
Q since it would hold Q in equilibrium. Hence, 

Q = as\ or si=—. 

a 


W lb. per ft. 


Q 3 

rnniiiiiiiiiiiiiiiiiiiiii. 

r Q 


& 

K *- 4 



(a) 


W lb. per ft. 


TTlllllfTTTIT 

A A 

— 

Pt^l' 

L i J 

lH 2 *1 

B /B / 
as t =Q_ 
(6) 


Fig. 159. —Stress in beam subjected to end-loads; deflection neglected. 


If, on the other hand, the bending loads were the only forces acting, 
the bending moment M at the mid-section (equal to \wl 2 , Fig. 92) 

would develop the resisting moment S -~- as shown in Fig. 159(6), 

the fiber unit-stress s 2 varying directly as the distance of the fiber 
from the neutral axis provided that its value does not exceed the 
proportional limit of the material. 1 And since the resisting 
moment holds the bending moment in equilibrium, the two 
moments are numerically equal. Hence, 


s 2 I Me 

M = — or s 2 = —• 
c I 


The normal unit-stress at any point of the mid-section then, 
according to the principle of superposition, is the algebraic sum of 
the unit-stresses caused by the loads acting separately; the max¬ 
imum unit-stress is a compressive stress and occurs on the top 
fiber; its value is 

S = Si + S 2 , 

_Q,Mc 
a' I ' 

1 For other limitations and assumptions see Art. 43. 
















186 


COMBINED AXIAL AND BENDING LOADS 


The unit-stress on the bottom fiber may be either tensile or com¬ 
pressive according as S 2 is larger or smaller than s\. As shown in 
Fig. 159(6), $2 is larger than si, therefore, the unit-stress on the 
bottom fiber is a tensile stress, and the surface of zero fiber-stress 
is some distance below the centroidal plane of the beam. Further, 
if the load Q were a tensile load the maximum unit-stress would 
occur on the bottom fiber and would be a tensile stress. 

Deflection of Beam Not Negligible .—If the deflection of the 
beam is not negligible, the load Q (Fig. 160a) cannot be considered 
to be an axial load with respect to the mid-section (or any other 
cross-section except the end sections, see Art. 3 and Fig. 2). Let 
A denote the deflection of the axis of the beam at the section on 
which the stress is to be found (in this problem the mid-section). 


'll) IK nor ft 



Q 


(a) 


Fig. 160.—Stress in beam subjected to end loads; deflection considered. 

Now if two equal opposite and collinear forces Qi, Q 2 (each equal 
to Q), are applied to the beam as shown in Fig. 160(6), the force Q 
will be resolved into a force Qi, which is axial with respect to the 
mid-section, and a couple having a moment QA; these two forces 
do not change the stresses developed on the mid-section, but 
simply modifies the force system so that the equations (P = as 
sl\ 

and M=—\ already developed in the preceding chapters can be 

made to apply to this problem. Thus, since Qi passes through the 
centroid of the mid-section, the unit-stress si due to Qi alone is 
constant over the section and is obtained from the equation 


Qi = Q = asi, 


as indicated in Fig. 160(6). The other forces constitute a bending 
moment composed of the cross-bending moment M, as in the pre¬ 
vious problem, and the moment QA. The total bending moment 
is therefore, Af+QA and is held in equilibrium by the resisting 


moment —. Therefore, 
c ’ 


















ILLUSTRATIVE PROBLEM 


187 


as indicated in Fig. 160 ( b ). Thus the maximum unit-stress s is, 
S = Sl + S2, 

= ^+(M+QA)|.(131) 

But since the value of A depends on the value of the total 
bending moment (M-\-QA) and the value of the bending moment 
depends in turn on the value of A, a method of approximation in 
solving equation (131) is, as a rule, the most convenient method. 
Thus, the value of A that would be caused by the cross-bending 

moment, M, alone is found first (equal in this problem to 

384 El 

Art. 46); this value of A is then used in the expression M-\-QA 
and this new value of the total ending moment is used to find a 
closer approximation to the value of A. This operation may be 
repeated as many times as desired. 

ILLUSTRATIVE PROBLEM 

Problem 153. —In Fig. 161(a) is shown a wall bracket, the horizontal 
boom BC being a 7-in. 15-lb. I-beam. The beam is pin-connected to the post 



BE at B and to the rod AC at C. If the load P is 4900 lb. find the maximum 
fiber unit-stress in the I-beam, assuming that the weight of the beam is 
negligible. 

Solution.—Deflection Neglected .—The forces acting on the I-beam are 
shown in Fig. 161(6). The normal stresses at the mid-section due to the 
end;load B x and the cross-bending loads are shown in Fig. 161(c). The value 
of B x and of C x is equal to the horizontal component (^-component) of the 
























188 


COMBINED AXIAL AND BENDING LOADS 


tension in the tie rod AC, and this tension may be found by applying the con¬ 
ditions of equilibrium either algebraically or graphically to the forces acting 
on the pin at C or on the post DE; it is assumed that the student is famil¬ 
iar with methods of finding the forces acting on the members of pin-connected 
structures. The value of the tension in the rod is found to be 4600 lb. and its 
^-component is 3920 lb. Therefore B x equals 3920 lb. Further, from a steel 
handbook the area a of the cross-section of the I-beam is found to be 4.42 
sq. in. and the section modulus I/c is equal to 10.4 in.? 

As indicated in Fig. 161(c) the compressive unit-stress, s, at the top of the 
I-beam is 

$ = Si+S2 

Bx PI c 
~ a + 4 I 

3920 4900X8X12 1 

~4.42 + 4 X 10.4 

= 888+11,300 
= 12,200 lb. per sq. in. 

Deflection Not Neglected .—The deflection A caused by the transverse forces 
alone is (Art. 47) 

_J_PZ_ 3 _ 1 4900(8-12) 3 

“ 48 ~EI ~ 48 30,000,000 -36.4 

= 0.0825 in. 

When the beam deflects, the load B x has a moment B X A with respect to 
the mid-section and, as explained above, the bending moment is M-\-B x A. 
Thus the compressive unit-stress on the top fiber of the beam is 


S = Si+S 2 , 



1 

= 888+11,300+4900 • 0.0825 • 

= 888+11,300+31.2'. 

Thus, very little error is introduced in this problem by neglecting the deflec¬ 
tion of the beam. 

87. Eccentric Longitudinal Load in Plane of Symmetry.— 

In Fig. 162(a) is represented a short 2 compression member acted 

2 The effect of an eccentric load on a long compression member (column) 
in which the deflection of the member must be considered, is discussed in 
Art. 104. 




ECCENTRIC LONGITUDINAL LOAD 


189 


on by an eccentric load P in a plane containing an axis of sym¬ 
metry 3 of each cross-section of the member, the amount of the 
eccentricity being denoted by e. Let it be required to find the 
maximum unit-stress developed on any normal cross-section, a, 
of the member. 

The force P may be resolved into an axial load Pi (equal to P) 
and a couple having a moment Pe (or P 2 e since Pi = P 2 = P) with¬ 
out affecting the stresses developed on the section (Fig. 162t). 



Fig. 162. —Stress due to eccentric load in plane of symmetry. 


Now if Pi were acting alone it would cause a constant unit-stress 
si on the area a such that, 

Pi = P = asi. 

And, if the forces P and P 2 were acting alone their moment Pe 
would cause the resisting moment (Fig. 1616) such that 



3 The stress caused by an eccentric load that does not act at a point on an 
axis of symmetry or principal axis, is discussed in Art. 88. All axes, however, 
are principal axes for a member having a square or circular cross-section; 
that is, the moments of inertia with respect to all axes in the plane of the area 
passing through the centroid of a square or circular area are equal. 




















190 


COMBINED AXIAL AND BENDING LOADS 


The combined effect of the axial force and couple is to produce a 
maximum compressive unit-stress $ such that 

S = Si + S2 

PMc 
a I 

_P .Pec 

~a~T 


and the minimum stress is —— j-. Or, since I may be expressed 

by the equation I— ak 2 in which k is the radius of gyration of the 
area a with respect to the centroidal-axis, the unit-stress at any 
point whose distance from the centroidal axis is c may be found 
from the equation 



in which s must not exceed the proportional limit of the material. 

Limitation on Eccentricity to Prevent Tensile Stress. —If the 
unit-stress S 2 at B (Fig. 1626) due to the moment Pe is greater 
than the unit-stress Si due to the axial load Pi, the resulting 
normal unit-stress at B will be a tensile stress. If, then, a tensile 
stress is to be avoided, as is usually desired in brittle material 
such as concrete, brick, cast iron, etc., which are relatively weak 
in tension, the value of e should not be greater than that found by 
equating S 2 equal to s\. Thus, 

Pec = P 
I a’ 
or 

Pec = P 
ak 2 a’ . 


6 c * 


Therefore, a short compression member will not be subjected 
to a tensile stress if the eccentric load acts on an axis of symmetry 

k 2 

of the cross-section at a distance not greater than — from the cen- 

c 

tral axis of the member. 


ILLUSTRATIVE PROBLEM 


191 


If the member has a rectangular cross-section the value — is 

c 

ib or where b and h are the dimensions of the cross-section. 
This fact is expressed by the common rule that in the design of 
masonry structures the load should not lie outside the middle- 
third of the central axes of a rectangular cross-section. If the 

k 2 

cross-sectional area of the member is circular the value of — is 

c 

\d, where d is the diameter, and hence the load should not lie out¬ 
side the middle-fourth of any diameter of the section if tensile 
stress in the member is to be avoided. 

(In each of the following problems the load lies in a plane of 
symmetry of the cross-section of the area on which the stress is to 
be found.) 

ILLUSTRATIVE PROBLEM 

Problem 154.—The machine member shown in Fig. 163(a) is acted on by 
a force P of 3000 lb. Find the maximum normal unit-stress on the area of 
the section AB at the wall. The dimensions of the cross-section of the bar 
at section AB is f in. by 3 in. 



Solution. The load P may be resolved in two components, a cross-bending 
load P v and a longitudinal eccentric load P x (Fig. 1636); and P x may be 
resolved further into an axial load P" and a couple having a moment P x e , 
by introducing the two equal opposite and collinear forces P' and P", each 
equal to P x , as indicated in Fig. 163(6). 

Now since P" is an axial load with respect to the section AB, it would, if 
acting alone, cause, and be held in equilibrium by, a total stress as i, the unit- 
stress Si being constant over the area a of the section as indicated in Fig. 
163(6). And the couple P x e, if acting alone, would develop a resisting moment 

— equal to the moment (P x e) of the couple as shown in Fig. 163(6). Again 
c 

s 3 J 

the cross-bending load P v would develop a resisting moment — equal to 

the moment ( P v l ). The shearing stress on the area is neglected in this dis¬ 
cussion. 
























192 


COMBINED AXIAL AND BENDING LOADS 


The normal unit-stress on the top fiber at A, then, is 

= —Si —S2+S3 

P cos 30° P cos 30°-1.5-1.5 P sin 30° •12-1.5 
2 I 5 ST® 5 + tV i-(3) 8 

2600 2600-2.25 , 1500-18 
~~ ~ 2.25 L69 1 1.69 

= -1150-3460 + 16,000 
= 11,390 lb. per sq. in., tensile stress. 

And at B the unit-stress is, 

Sb= —S1+S2—S3 

= -1150+3460-16,000 
= —13,690 lb. per sq. in., compressive stress. 

PROBLEMS 

155. In Fig. 164 is shown a machine member having a rectangular cross- 
section 1 in. by 4 in. It is acted on by a force P 2 of 18,000 lb. and a force Pi, 
the action lines of which are shown in the figure. Find the value of Pi if 
the maximum tensile unit-stress is 20,000 lb. per sq. in. Ans. P 1 = 8270 lb. 



156. What value should e, in Fig. 165 have in order that the unit-stress 
at the top fiber within the middle third of the beam, due to the load Q shall be 
equal (and opposite) to the unit-stress due to the cross-bending loads; Q = 9P 




157. The frame shown in Fig. 166 is used for small riveting, punching and 


















































PROBLEMS 


193 


stamping machines. Find the unit-stress developed at A and at B when the 
load P is 2000 lb. 

Ans. s^ = 1470 lb. per sq. in., tension; sb = 1870 lb. per sq. in., compression- 

158. The inclined beam shown in Fig. 167 carries a load P of 5000 lb. at 
the mid-span. If the length, l, is 10 ft., what is the maximum compressive 
unit-stress developed in the beam? 

Ans. $ = 1080 lb. per sq. in. 


159. The post DE of the wall bracket shown in Fig. 161(a) is made of 
two 6-in. 15.5-lb. channel sections latticed together, the 6-in. dimensions being 
parallel to the direction of the boom BC. Find the maximum compressive 
unit-stress on a section of the post just beneath the section passing through A. 
Assume the load to be at the outer end of the boom BC. 

160. A cast-iron machine frame shown in Fig. 168 is subjected to a load P 
of 8000 lb. The area of the cross-section at AB is 40 sq. in. and the centroidal 
axis YY is 6 in. from the outer edge of the section. The moment of inertia 
of the area with respect to the centroidal axis is 400 in 4 . Find the maximum 
tensile and compressive unit-stresses on the section AB. 

Ans. St = 1960 lb. per sq. in., tension; s c = 2440 lb. per sq. in., compression. 




161. The small crane shown in Fig. 169 has a clear swing of 28 in. Find 
the load P which will cause a maximum compressive unit-stress of 9000 lb. 


per sq. in. on the inner edge at B. 


162. The open link shown in Fig. 170 is made 
of a steel bar having a diameter of 2 in. If the 
tensile unit-stress at A is 18,000 lb. per sq. in. what 
is the value of the load P? 


Ans. P = 1090 lb. 






































194 


COMBINED AXIAL AND BENDING LOADS 



163. A timber post having a cross- 
section 6 in. by 6 in. supports one end 
of an inclined beam as shown in Fig. 
171. The load P is 4000 lb. Find the 
unit-stress at points A and B of the 
section AB of the post. 


88 . Eccentric Load Not in Plane of Symmetry.—In the pre¬ 
ceding article the action line of the resultant longitudinal load 
passed through a point on an axis of symmetry of the area on which 
the stress was desired. Now, as stated in Art. 43, in using the 
si 

flexure formula M =— the neutral axis is assumed to be perpen- 
c 

dicular to the plane of the loads, but this is true only when the loads 
lie in a plane of symmetry . 4 

If the load P (Fig. 172a) does not lie in a plane containing an 
axis of symmetry it may, as in the preceding article, be resolved 
into an axial load P i (equal to P) and a couple, P and P 2 , having 
a moment Pe, and this couple may be further resolved into two 
component couples in planes containing the axes of symmetry or 
principal axes, OX and OF; the stress due to each component 
couple may then be found from the ordinary flexure formula. 
Thus, the moment of the component couple in the plane contain¬ 
ing the axis OX is Pe cos 8 or Pe x (Fig. 172a), the forces of this 
couple being represented by P 2 and P 3 in Fig. 172(b), and the 
moment of the component couple in the plane containing OF is 
Pe y and the forces of this couple are P 4 and P 5 . These five 
forces, each equal to P, will produce the same unit-stress at any 
point in a section, such as the section ABCD, as the original force 

4 The more general statement is that the loads must lie in a plane contain¬ 
ing a principal axis of inertia of the cross-section; an axis of symmetry is 
always a principal axis; however, a section always has a principal axis even 
though it may not have an axis of symmetry. 











ECCENTRIC LOAD NOT IN PLANE OF SYMMETRY 195 


P, and the normal unit-stress at any point in the section is the 
algebraic sum of the unit-stresses produced by the axial load Pi 
and the two bending couples Pz, Ps, and P 4 , P 5 . 



Fig. 172.—Eccentric load not in plane of symmetry. 


The maximum unit-stress occurs at C and is 

S = Si + S2-fS3 


P I PC%Cx I PCyCy 


For other points in the area a either the second or the third 
term or both may be negative. Thus the unit-stress at a point E 
in the quadrant O'A (Fig. 1725) is 


s = .(132) 

Cl Iy lx 

Kern of a Section .—It is evident that the least value of s will 
occur at A and its value will be given by equation (132) when 

7 7. 

x = - and y= 7 :, and it will be a tensile stress if the sum of the 
2 2 

p 

last two terms is greater than —. If a tensile stress is to be avoided 

therefore, the values of e x and e y can not be greater than those 
found from the equation 



P 
a’ 






















196 


COMBINED AXIAL AND BENDING LOADS 


or 


whence. 


Sy 2 _ 1 
bh’ 

@x i 1 

6+F=6 ° r 


+ ^ ’ 


which is the equation of a straight line that intersects the axis 

OX at a distance ^ from 0. and the axis OY at a distance § from 0 
6 6 

(Fig. 173). Similar limits occur in the other quadrants and hence 

the resultant load on the member must 

pass within the shaded area shown in 

Fig. 173 if tensile stress in the member 

is to be avoided. 

The area within which the resultant 
load must pass to avoid tensile stress on 
a section is frequently called the core, 
kernel, or kern of the section. Thus for 
a rectangular section the kern is a rhombus the diagonals of 
which are the middle-thirds of the principal axes of the section. 



<- 1 

) - » 



a°4) 

- 

O* 



Fig. 173.—Kern of a 
section. 


PROBLEMS 

164. A short rectangular timber having a section 6 in. by 8 in. is subjected 
to a longitudinal eccentric compressive load of 10,000 lb. The action line 
of the load passes through a point in each section 3 in. from the 6 in. side and 
2\ in. from the 8-in. side. Find the unit-stress at each corner. 

Ans. s A = 56 lb. per sq. in. T., Sjg = 156 lb. per sq. in. C. 

5(7 = 468 lb. per sq. in. C., sd = 264 lb. per sq. in. C. 

165. The resultant normal pressure, P, on top of a concrete base having a 
square cross-section, a, acts at the center of one quadrant of the square. Will 
tensile stresses occur in the base? If so, find the maximum tensile and com¬ 
pressive unit-stress that will be developed, assuming the values of P and a to 
be 140,000 lb. and 4 sq. ft., respectively. 

89. Eccentric Loads on Riveted Connections. —Examples of 
eccentric shearing loads are found frequently in riveted joints. 
Whenever feasible, however, eccentric loads should be avoided; 
that is, the action line of the resultant force to be transmitted 
through the rivets should pass through the centroid of the total 
rivet shearing area. Care in securing this condition, however, fre- 









ECCENTRIC LOADS ON RIVETED CONNECTIONS 197 


quently is not given the attention it merits. The shearing stresses 
in riveted joints subjected to eccentric loads may be found as 
follows. 

Let it be required to find the shearing unit-stress in each 
rivet in the riveted joint shown in Fig. 174(a). A force P having 
the eccentricity e is transmitted from the member A to the gusset 
plate B, and from the gusset plate through the four rivets to the 
member C. The shearing area of each rivet will be denoted by a, 
the diameter by d, and the distances of the centers of rivet areas 
from the centroid, G, of the total shearing area by h, I 2 , etc. 




Fig. 174. —Shearing stress in rivets due to eccentric load. 

The load P may be resolved into a load P\ (Fig. 1746) acting 
through the centroid of the total shearing area and a twisting or 
torsional couple Pi, P 2 having the moment Pe. The central shear¬ 
ing load Pi if acting alone would develop the same shearing unit- 
stress, s' s , on each of the rivet areas (or rather this is the assump¬ 
tion usually made, see Art. 13). 

The resisting stress that holds Pi in equilibrium, then, is 4a s' 8 . 
Thus, 

p 

Pi = 4as' s or s' s = —, 

4a 


as indicated in Fig. 174(6). 

And, if the couple Pe were acting alone it would develop a 
/s" J \ 

resisting moment (——, Art. 26 ), 5 the shearing unit-stress in the 


5 In Art. 26 it was stated that — is the expression for the resisting moment 
c 

only for a solid or hollow cylindrical shaft. The rivet areas, however, may 
be transformed into annular areas having radii l\ and 1 2 , similar to the cross- 
section of two concentric hollow cylindrical shafts. 






















198 


COMBINED AXIAL AND BENDING LOADS 


rivets varying directly as the distances of the rivets from the cen¬ 
troid G, as shown in Fig. 174(6). This resisting moment holds 
the moment Pe in equilibrium. Hence, 



Pec 

~T' 


The shearing unit-stress on the bottom rivet, then, is 

Ss = S 8 + 5 S 


P Pec 
a J~’ 


(133) 


in which c is the distance from G to the rivet on which the stress 
is desired (equal to l 2 for the bottom rivet), and J is the polar 
moment of inertia of the total shearing area of the rivets with 
respect to an axis passing through G. 

Since the diameters of the rivet areas are small in comparison 
with the distances of the rivets from G, the value of J is, with 
slight error, the sum of the products obtained by multiplying 
each rivet area by the square of the distance of its center from G. 
Thus, 

J = 2a(h 2 +l 2 2 ) 

and hence 

Pg = ZO£_ s .( 134 ) 

12 


Instead of obtaining the resisting moment from the expres¬ 
s'^/ 

sion ——, it may also be found as the sum of the moments, with 
c 

respect to G, of the total shearing stresses on the rivet areas. Thus 
if s" s denotes the shearing unit-stress on the rivets farthest from G, 
and s" 8l on the rivets nearest to G, we have 


Pe = 2[as" sh+as' s ih] 

= 2^as // s ?2+a^s // s ^i"|, 


• / G // 

since s si = t s s 

*2 


=2 

12 


which is the same as equation (134). 



PROBLEMS 


199 


PROBLEMS 


166. The diameter of the rivets in the joint shown in Fig. 175 is f in., 
and the load P is 4000 lb. Find the shearing unit-stress in the top and bottom 
rivets. Ans. s = 6750 lb. per sq. in. 




167. The load P transmitted through the joint shown in Fig. 176 is 5000 lb. 
and the diameter of the rivets is f in. Find the shearing unit-stress in the 
top and in the bottom rivet. 



168. Fig. 177(a) shows a commonly used joint in a riveted Pratt truss. If 
the joint is arranged as shown in Fig. 177(6) how much will the maximum 
shearing unit-stress in the rivets in the top chord A be decreased? All the 
rivets have a diameter of f in. The values of Pi and P 2 are such as to produce 
a shearing unit-stress of 10,000 lb. per sq. in. in the rivets connecting the two 
members, B and C. to the gusset plate. (See Prob. 38.) 

90. Helical Spring. Stress Developed .—A helical or coil 
spring (Fig. 178a) is frequently used to resist an axial load P that 
lengthens or shortens the spring. The stress developed in the 
rod of which the spring is made is mainly torsional shearing stress* 
















































200 


COMBINED AXIAL AND BENDING LOADS 


The value of the shearing unit-stress in a spring made of rod having 
a circular cross-section may be found as follows: 

Let a section be passed through the rod at A (Fig. 178a) and 
the lower part of the spring be removed. The part above the sec¬ 
tion A is shown in Fig. 178(6). Further, let the load P be resolved 
into a force Pi (whose action line passes through the center of the 
section at A) and a twisting moment Pe, by introducing the tw T o 
forces Pi and P 2 (Fig. 1786) each equal to P. Now these forces 



(a) 



Fig. 178.—Stress in helical spring. 


(which produce the same stress on the section at A as does the 
original one force P) are held in equilibrium by the stresses on the 
section at A. If the force Pi were acting alone it would produce 
a resisting shearing stress that would hold Pi in equilibrium, and 
this direct shearing stress would be equal to as' 8 , as indicated in 
Fig. 178(6), a being the area of cross section of the rod, and s' 8 
the shearing unit-stress. Hence, 

Pi = P = as' 8 or 


Further, if the twisting moment Pe due to the couple P, P 2 were 

s" J 

acting alone it would produce a resisting moment —on the area 

a which would hold the external or twisting moment in equilibrium 
as indicated in Fig. 178(6). Hence 

D s" 8 J 
Pe = -. 









HELICAL SPRING 


201 


Now the direct shearing unit-stress s' 8 in most springs is negli¬ 
gible in comparison with the torsional unit-stress s" 8 , and hence 
the chief unit-stress in the rod is due to torsion and its maximum 
value is 

„ _Pec 

* s— j , 


in which e is the mean radius of the coil, c the radius of the rod 
^equal to ^ where d is the diameter of the rod^ and J is the polar 

moment of inertia of the cross-sectional area of the rod 
Art. 162). 

The above expression then may be written 

Pe- 

„ _ 2_16Pe 

s 7 rd 4 7 rd s 

32 


' j _ 7T d 4 

“ 32 ’ 


If, however, the direct unit-stress s' 8 is not negligible, the 
maximum unit-stress on the area a will be the sum of s' s and s" 8 . 
Thus 

P , 16Pe 

S s = —h— 
a ird 6 



(135) 


Deflection of Helical Spring .—The increase or decrease in 
length (called deflection) of the spring due to a given load P may 
be found as follows: 

If the shearing proportional limit of the material is not exceeded, 
the work done by the load, as it increases uniformly from zero to 
the value P, is stored in the material as stress (potential) energy, 
and further, if the stress developed is assumed to be due only to 
torsion, it follows that 

Work done in causing deflection = Work done in twisting the rod. 


Now the work done by the load in deflecting the spring is 
the product of the average value of the load j and the deflec¬ 
tion A. Similarly, the work done in twisting the rod through an 





202 


COMBINED AXIAL AND BENDING LOAD 


angle 6 is the product of the average value of the twisting moment 
/ T Pe\ 

(or — j and the angle of twist. Hence, 
iPA = iTe = iPed. 

A = ed. 


Therefore, 


But from Art. 28, the value of 6 , is given by the equation, 

Pel 


6- 


EJ' 


Further, for a closely coiled spring, l = 2iren, approximately, where 
n is the number of turns. Hence, 


A — e 


Pe2iren 

~EJ~’ 


7rd 4 


but, since the above expression becomes, 


A = 


64Pe 3 n 

~E~dF' 


(136) 


(137) 


If A is expressed in inches and P in pounds, then e and d are 
expressed in inches, and E s in pounds per square inch. If the 
shearing unit-stress developed in the bar is greater than the shear¬ 
ing proportional limit the above equation is not valid. 

PI 

The deflection, A s , due to the direct shear is A s =—^and hence 

dHj 8 

the total deflection is 


PI . Pe 2 l PI ( 


aE, ' EJ aE s { 


total deflection = —pH— 

an § Hj 8 fJ 

in which k is the polar radius of gyration of the section. 


(138) 


PROBLEMS 

169. A helical spring when compressed 4 in. must support an axial load of 
1600 lb. The diameter of the rod is f in., the mean diameter of the coil is 
3 in. and the working shearing unit-stress of the material (spring steel), is60,000 
lb. per sq. in. Find the number of coils required. Is the working stress 
exceeded? Ans. n = 21.2 coils. = 50,000 lb. per sq. in. 






PROBLEMS 


203 


170. A helical spring made of steel rod \ in. in diameter with 5 turns is 
2.65 in. high and has an outside diameter of 1.35 in. Find the modulus of 
the spring, that is, the force required to stretch it an inch, assuming that the 
proportional limit of the material is not exceeded. 

Ans. Mod. = 965 lb., if deflection due to direct shear is neglected. 

171. A bronze helical spring is required to support 980 lb. when deflected 3 
in. Find a suitable diameter of the coil and number of turns if the diameter 
of the rod is £ in. and the working unit-stress not to exceed 50,000 lb. per sq. 
in. E s = 5,000,000 lb. per sq. in. 


CHAPTER XI 


COMPRESSION MEMBERS. COLUMNS 

91. Introduction. —A compression member is a bar (either 
solid or built-up, and, as a rule, approximately straight) subjected 
to an end-load (or loads) acting parallel to the axis of the bar. 
If the load acts through the centroid of the end section it is called 
an axial load; if not, it is called an eccentric load. But, an eccen¬ 
tric load is equivalent to an axial load and an end-moment or 
couple. The character of end-loads applied to compression 
members as used in engineering structures and machines depends 
largely on the type of end-connection used in the structure, and 
on the relative stiffness of the members connected. For example, a 
locomotive connecting rod, many members in certain types of 
bridges, etc., are connected by pins, and the end-load exerted on a 
compression member by the pin is mainly a single force, since the 
end-moment due to the friction of the pin is relatively small; 
whereas, most members in buildings, etc., are connected by rivets 
either directly or by means of an auxiliary piece such as a gusset 
plate, and a compression member may be subjected to an end- 
moment, in addition to a compressive load, due to the bending 
of a member (a beam, say) to which the compression member is 
attached. 

If a compression member is slender (relatively long), it is 
usually called a column; and if relatively short it is called a com¬ 
pression block or a strut. This chapter deals mainly with the 
strength of columns; there is, however, no definite slenderness 
that differentiates a column from a compression block. 

In the investigation of the stresses and deformations in a 
column there must be considered, as affecting the action of the 
column, certain conditions that may be neglected in the investiga¬ 
tion of the stresses and deformations in most other structural 
members. For example, in determining the stresses and deforma¬ 
tions in beams, tension members, torsional shafts, etc., when 

204 


DISTINGUISHING FEATURES OF COLUMN ACTION 205 


resisting loads, the member in question is considered to satisfy 
ideal conditions; namely, the body is considered to be straight, 
to be made of homogeneous material free from initial stresses, to 
be so constructed that it acts as a whole or as a unit, and to 
be subjected to loads having definitely known action lines. And, 
although it is known that such ideal conditions are never fully 
realized in these structural members, experiments and experience 
have shown that, in general, the results found from an analysis 
of the actions in the ideal beam, tension member, etc., are reliable; 
that is, the primary or significant action in the actual member is 
the same as that in the ideal body. But, the action in an actual 
column, as shown by experiments and experience, may deviate 
much from that in an ideal column. This fact may be stated in 
a somewhat different way as follows: In beams, tension members, 
etc., conditions that differ from those in the ideal members do not 
have more than a secondary effect on the stresses and deforma¬ 
tions in the beams, tension members, etc., whereas the similar 
conditions in a column may be of prime importance in determining 
the resistance of the column. 

Further, a quantitative measure of the effects of these con¬ 
ditions cannot, in general, be determined, and hence they render 
the analysis of column action less definite and satisfactory than 
that of beams, tension members, etc., and at the same time, cause 
a column to be a more critical member in a structure than most 
other structural members. 

A brief discussion of the actions and conditions that must be 
especially considered in a column, are given in the next article and 
also in Art. 97. 

92. Distinguishing Features of Column Action. —The funda¬ 
mental idea of column action is that of combined compression and 
bending, but the bending action in a column is accompanied by 
certain conditions that renders a quantitative measure of the bend¬ 
ing action less certain than that of the bending in a beam or in a 
member subjected to combined tension and bending, etc. The 
distinguishing features of column action may be stated briefly 
as follows: 

1 . A straight column may bend or deflect as a whole when sub¬ 
jected to a supposedly axial end-load (Fig. 179a), the initial bend¬ 
ing being due to a slight initial or accidental eccentricity] 1 of load- 

1 If the column bends there must be a bending moment to start the bend- 


206 


COMPRESSION MEMBERS, COLUMNS 


ing, and this deflection causes an increase in the moment-arm of 
the load, and hence increases the bending moment; the increased 
bending moment in turn increases the deflection, etc., tending to 
cause failure of the column by flexure or what is frequently called 
“ buckling.” This action is of special importance in the more 
slender columns and in the component parts of built-up columns. 

In a column, then, the bending moment may increase more 
rapidly than the load increases, whereas this undesirable condition 
does not occur in other structural members. For example, in a 
straight compression block the deflection may be assumed to be 
negligible and hence the initial bending moment, if any, due to 
initial eccentricity, etc., increases only as the load increases since 
the moment-arm of the load remains constant. Likewise, in the 
case of a beam, although the deflection of the beam may be greater 
than that of a column, the deflection has a negligible effect on the 
moment-arm of the load. Again, in the case of a tension mem¬ 
ber subjected to an eccentric load, the bending of the member 
causes the moment-arm of the load to decrease, and hence the 
bending moment increases even less rapidly than the load increases. 

2. By virtue of the action discussed under (1) above, an initial 
crookedness in a column as a whole (and all columns are more or 
less crooked) reduces the strength of a column more than a like 
crookedness reduces the strength of a beam, tension member, etc. 

3. Local kinks or bends in the component parts ot a built-up 
column, lack of homogeneity of material, initial stresses 2 in the 
material, and poorly designed end-connections, may cause stresses 
and deformations in parts of a column greatly in excess of those 
assumed to exist in the corresponding ideal column, and these 
localized stresses and deformations, as a rule, reduce the resistance 
of the column more than similar localized stresses reduce the 
resistance of beams, tension member, etc. 

4. A variation in the position of the action line of a load on a 
column from that assumed in the ideal column, that is, the pres¬ 
ence of accidental or fortuitous eccentricities, cause, in general, 
a much greater increase in the stresses in a column than similar 
variations cause in the stresses in beams, in tension members, etc. 

ing; various conditions contributing to this bending are discussed later but an 
initial or accidental eccentricity may here be considered to be the chief cause 
of the initial bending. 

2 For causes of initial stresses see Art. 106 and 137. 


SLENDERNESS RATIO 


207 


Further, such fortuitous eccentricities are very likely to occur due 
to the actions of the end-connections. 

93. Slenderness Ratio. —The slenderness ratio for a column is 
a term that enters in all column formulas; it is the ratio of the 
length, l , of the column to the radius of gyration, r, of the cross- 
section with respect to the neutral axis, the neutral axis being the 
centroidal axis that is perpendicular to the plane in which the 
column bends or tends to bend. If the column is pivot-ended, 
that is, free to turn at its ends, it will bend in a plane perpendicular 
to the axis of the cross-section about which the moment of inertia 
is least, and r will be the least radius of gyration. Column sec¬ 
tions are frequently designed so that the moments of inertia of 
the section about the principal, axes are approximately equal; 
this condition, however, is not always desirable. 

94. Two Limiting Cases of Compression Members.— If a 
short solid compression block is subjected to an end load having a 
very slight eccentricity, the block may be assumed to be in pure 
compression since the bending action may be neglected. But if 
this block be assumed gradually to become longer (more slender) 
the tendency to bend becomes greater, and the bending cannot be 
considered to be negligible if the length of the member is greater 
than about ten times the least lateral dimension; that is, if the 
slenderness ratio is about 40. And, if the column is very slender, 
the resistance of the column is mainly its resistance to bending, 
the effect of the direct compression then being negligible. The 
slenderness ratios of compression members used in most engineer¬ 
ing structures and machines, however, are such that both direct 
compression and bending are of importance, and such members 
will be discussed after considering the two limiting cases of com¬ 
pression members. 

One limiting case of a compression member, then, is a com¬ 
pression block, and for such a member, when made of homo¬ 
geneous material and subjected to an axial load P, the formula is 


P = as, 


in which a is the cross-sectional area of the column and $ is the 
unit-stress at all 3 points in the area. 

3 The variation of stresses over the cross-section in short built-up columns 
is discussed in Art. 106 and 137. 


208 


COMPRESSION MEMBERS, COLUMNS 


The other limiting case is that of a member so slender that 
bending or flexure is the only action that need be considered, and 

si 

hence the flexure formula, M = —, for such a column applies, the 
bending moment M at any section being Py. (Fig. 179a). Thus, 


Py 


si 


which states that the bending moment (Py) of the external forces 
about any section is held in equilibrium by the resisting moment 

of the stresses developed on that section. But the value of 


(?) 


■Py 


the deflection y in the equation corresponding to a given load P 
is unknown and hence the value of P that 
produces a given stress s cannot be found, since 
there are two unknown quantities in the equa¬ 
tion; in other words, the problem is statically 
indeterminate; but, as discussed in Chapter 
VII, the elastic curve equation offers another 
method for finding external forces. The expres¬ 
sion for the load that causes a slender column 
to bend as found from the elastic curve of 
the column was first obtained by Euler in 1757 
and is known as Euler’s column formula; it 
is derived in the following article. 

95. Euler’s Column Formula.—The prob¬ 
lem is to find the least value of an axial load 
P that will cause a column to bend. The im¬ 
portance of this value of the load lies in the 
fact that, due to the conditions stated in Art. 
92, a column fails by bending if the load is increased only a small 
amount above this value, and hence the least value of the load 
that causes a column to bend is also the maximum load the 
column can resist, unless the column fails by crushing of the 
material by direct compression before this value is reached. 

Now the method of obtaining this critical bending load is to 
assume that an ideal column is deflected by a lateral force 4 and 
that an axial load P is applied that will maintain this deflection 



Bending 
of slender column. 


4 An axial load could not cause an ideal column to bend. 








EULER’S COLUMN FORMULA 


209 


after the lateral force is removed; a relation between the load and 
the deflection is then found from the elastic curve equation of the 
deflected column, and the limiting value of P found from the equa¬ 
tion, by expressing the fact that the deflection is indefinitely small, 
is the least value of P that will cause an ideal 
column to bend (or the maximum value that 
will not cause bending). 

In Fig. 180(a) is represented a slender 
solid column of constant cross-section; it is 
assumed to be straight, to be made of homo¬ 
geneous material free from initial stresses, 
and to be subjected to an axial end-load only, 
that is, no restraints occur at the ends and 
the weight of the column is neglected. Let 
the load P be such as to hold the column 
in the deflected position as shown in Fig. 

180(6). Now since bending is the only action 
here considered the equation of the elastic 
curve is the same as that of a beam; namely 
(see Art. 45), 



±EI W> =M ’ 


Fig. 180. — Elastic 
curve of slender 
column. 


in which M is positive and equal to Py, and, if the origin of axes 
is chosen at the upper end of the column and the positive direc¬ 
tion of the x and y axes are chosen downward and to the right, 

d 2 v 

respectively, as shown in Fig. 180(6), then is negative (see 
Art. 45), and the above equation becomes 

. (139) 

This equation may be written 
d\ 




EI- 


dx 


= -Py, 


(139 a) 


dy 


and if each side of the equation is multiplied by each side 

becomes an exact derivative of the form udu ; namely, 

Ti r fdy\ j/dy\ D , 











210 


COMPRESSION MEMBERS, COLUMNS 


Integrating, we have, 


®(D’— 


(140) 


dy 


Now simultaneous values of ~ and y are not known, and hence 

Ci cannot be expressed in terms of the known quantities E, I 
and l, but it may be expressed in terms of the maximum deflec- 
dy 

tion, A. Thus, since ^ = 0 when y = A, we have 


0=—PA 2 +Ci; whence Ci=PA 2 , 
and eq. (140) becomes 

EI {S) 2=p(A2 ~ y2) . (141) 

This equation may be integrated after the variables are separated; 
the equation may be written as follows: 


dy 


Va 2 


y —- l- p - dx 


Integrating, we have 


sin_ 1 A = Vl/ a:+C2 ’ 


or 


!-»■ 


(~\[m x+C2 )' 

But y = 0 when x = 0 and hence C 2 = 0. Therefore, 5 


|P 


l = sinyJ^jx or y = A sin 


(142) 


\fr- • • 


(143) 


5 If the origin of axes is chosen at the center of the column as in Fig. 181, 
eq. (143) becomes 


y = A cos 



This may be shown as follows: The above derivation as far as eq. (142) is 







EULER’S COLUMN FORMULA 


211 


Now if the deflection of a column is considered to decrease, the 
value of P required to maintain the deflection also decreases and 
attains its limiting (minimum) value when the deflection becomes 
indefinitely small, and since the column is then virtually straight, 
the vertical distance, x, between the ends of the column is equal 
to l. Therefore, the limiting value of P may be found by sub¬ 
stituting the values y = 0 when x = l in eq. 143. Whence, 


sin (w*) =0 ’ 

and hence the angle m ust be equal to tt or some integral 


multiple of 7r; as discussed in Art. 100, the value is tt for a column 
with pivoted ends and 47r for one with fixed ends, etc. Therefore, 
the minimum value of P for a column having pivoted ends is 



(144) 


An axial load P, then, less than that given by eq. (144) will not 
cause a column to bend, whereas a load greater than this will cause 
an actual column to bend, or will maintain an ideal column in a 
deflected position. And, although an ideal slender column may 
resist a somewhat larger load than this, the accompanying deflec- 


independent of the origin of axes. Now referring to Fig. 181 
and eq. (142), we have, y = A when z = 0, and hence 


sin _1 1=0+C 2 , whence C 2 =^ 


Therefore eq. (142) becomes 



or 


whence, 




Fig. 181. 







212 


COMPRESSION MEMBERS, COLUMNS 


tion becomes excessive 6 when P has a value only slightly greater 
than that given by eq. (144). Therefore, the value of P in eq. 
(144) is considered to be the maximum load an actual (structural) 
column can resist without failing by bending. It may fail, how¬ 
ever, by crushing before this value is reached, as will be discussed 
later. 

The fact that excessive deflections occur when P has a value 
close to the Euler load is clearly shown by testing a slender col¬ 
umn, such as a wooden airplane 
strut, and measuring the deflection. 
The load-deflection curve obtained 
is shown in Fig. 182; the deflection 
starts somewhat before the Euler 
load is reached due to slight eccen¬ 
tricities, etc., but at the Euler load 
the deflection increases rapidly to a 
value equal to A B with very slight 
increase of load, the stress in the 
column being less than the propor¬ 
tional limit until the deflection AB is 
exceeded. 

Since 7 in eq. (144) is equal to ar 2 
(Art. 163) where a is the area of cross-section and r is the radius 
of gyration of the cross-section about the neutral axis, Euler’s 
equation may be written 

. 046) 



Fig. 182.— Relation between 
load and deflection for slender 
column. 


in which - is the slenderness ratio of the column. 
r 

Eq. (144) shows that the only property of the material on which 
P depends is the stiffness of the material, E (see Art. 144 and 145); 
therefore, a column made of high-carbon or special alloy steel 
would begin to bend, and hence fail, at the same load as would 


6 This fact is not evident from the analysis of the column action here pre¬ 
sented; it is shown by the true equation of the elastic curve, that is, the equa¬ 
tion obtained when the assumption that the length along the axis is the same 
as the length along the curye (dx = dl) is not made. The truth of the state¬ 
ment, however, is clearly shown by experiments as indicated by the results 
represented in Fig. 182. 







ILLUSTRATIVE PROBLEM 


213 


one made of low-carbon steel, provided that the - were relatively 

large so that the columns would not fail by crushing before the 
Euler load is reached. 


ILLUSTRATIVE PROBLEM 


Problem 172. A column made of a 10-in., 35-lb. I-beam and two plates f 
in. by 11 in., as shown in Fig. 183, is 32 ft. long and is subjected, in a testing 
machine, to an axial load applied through spherical-sea ted bearings. Calcu¬ 
late the least radius of gyration, r, and the corresponding slenderness ratio, 
l 

—. Find the maximum load the column can resist, and also a working load 
r 

using a reduction factor (so-called factor of safety) of 3. 


Solution .—The moment of inertia of the cross-sectional area about the 
z-axis may be found as follows: Area of each plate, ai = 6.87 sq. in. A steel 
maker’s handbook gives: 


Hence, 


Area of the I-section = 10.29 sq. in. 
I* of the I-section = 146.4 in. 4 

l y of the I-section = 8.5 in. 4 



Fig. 183. 


Total area of cross- 

section, a = 24.03 sq. in. 

l x of the two plates = 2(/+ai<2 2 ), (see Art. 164) 

= 2 aid 2 approximately 
= 2*6.87 • (5.3) 2 
= 386 in. 4 . 

l x for whole area = 146.4+386 = 532.4 in. 4 
I y for whole area = 8.5+2(^M 3 ) 

= 8.5+2( 1 Vf) (11)> 

= 8.5+138.6 = 147.1 in. 4 

Since I y is smaller than I x the column will bend so that the y-axis is the 
neutral axis, that is, in a plane perpendicular to the y-axis. 

The least radius of gyration, r= —= x / 2.47 in. Therefore, 

\ a \24.03 


l 32X12 
r ~ 2.47 


= 155.5, 


and this value is sufficiently large, as discussed in the next article, to allow 
the column to bend before failing by direct compression. 







214 


COMPRESSION MEMBERS, COLUMNS 


Hence, 


,, . , , P **E 9.87X30,000,000 100Knl , 

Max. unit-load = — = — 7— =- _ -= 12,250 lb. per sq. in. 

a l l \ 2 (155.5) 2 


Working load = 


P = 12,250X24.03 = 294,200 lb. 
294,200 


= 98,100 lb. 


96. Graphical Representation of Formulas for Ideal Columns.— 

As already noted, the maximum axial load P that an ideal column 
having a very small slenderness ratio (a compression block) can 
resist depends on the compressive strength, s, of the material and 

p 

the cross-sectional area a the equation being — = s, whereas the 

maximum load that a slender column can resist depends on the 
stiffness, E (not the strength), of the material and on the slender¬ 
ness ratio in addition to the area a; the equation being 


P 7T 2 P 



Now the results of tests of axially loaded columns agree approx¬ 
imately with the calculated values from the above equations when 

the columns have small and large values, respectively, of X Thus, 

if a series of ideal columns were made of one kind of material but 
with different slenderness ratios, and were tested to failure by 


applying axial loads, the columns having small values of - (less 

than about 40) would fail by direct compression when the com¬ 
pressive unit-stress in the material reached the ultimate com¬ 
pressive strength of the material; the maximum useable com¬ 
pressive strength of ductile material in columns, however, is the 
yield point of the material since at this stress plastic flow of the 
material occurs and the column fails to perform its structural 
function even if total collapse does not result. But the columns 


having large values of - would fail by buckling when the unit-load 





ILLUSTRATIVE PROBLEM 


215 


— is slightly greater than that given by Euler’s formula; the unit- 

stress in the column when the buckling starts being less than the 
proportional limit of the material. 

p 

If, then, values of — that cause failure are plotted (Fig. 184) 

as ordinates and the corresponding values of - as abscissae the 
P 

column formula — = s is represented by the horizontal line AB, 

and Euler’s equation is represented by the curve CBD. 

An ideal column 

having an ^ greater 

than that represented 
by the abscissa OB' 

(Fig. 184) fails by bend¬ 
ing, and an ideal column 
having a slenderness 
ratio less than this crit¬ 
ical value fails by the 
yielding of the material 
in direct compression 
if the material is duc¬ 
tile, or by crushing or 
diagonal shearing of the 
material if the material 
is brittle. 

The graph representing the column formulas for ideal columns 
is, therefore, the curve ABD of Fig. 184; the part BC of the 
Euler curve has no physical meaning, since the column would fail 
by crushing before a load represented by an ordinate to BC could 
be applied. 



Fig. 184.—Graphical representation of ideal- 
column formulas. 


ILLUSTRATIVE PROBLEM 

Problem 173. Find the critical value of corresponding to OR' in Fig. 184, 

r 

for an axially-loaded pivot-ended ideal structural-steel column, 

Solution .—The yield point of structural steel is about 40,000 lb. per sq. in. 






216 


COMPRESSION MEMBERS, COLUMNS 


and # = 30,000,000 lb. per sq. in. Now — as found from the two column 

a 

formulas are equal when - has the value OB' (Fig. 184), and hence, 
r 


That is, 


whence, 


P 

~a~ S ~ [1 


2 E 


40,000 = 


7T 2 X 30,000,000 



Therefore, an ideal axially-loaded pivot-ended structural-steel column would 
fail by buckling if its slenderness ratio were greater than about 86. Tests 
of actual pin-ended steel columns, however, show that Euler’s equation is 
l 

applicable only when - is greater than about 120, for reasons discussed in 
r 

Art. 92 and in the following articles. 


PROBLEMS FOR ARTICLES 96 AND 96 

l 

174. Find the least value of — for an ideal pivot-ended oak column that 

r 

will fail by bending when subjected to an axial load. Assume the following 
values for the ultimate compressive strength s u , and modulus of elasticity: 
s m = 8000 lb. per sq. in., # = 1,500,000 lb per sq. in. 

175. A 3-in. by 3-in. by \ in. wrought iron angle 15 ft. long was tested by 
the Pencoyd Steel Co., spherical seated bearings at the ends being used. It 
failed when subjected to an axial unit-load of 2650 lb. per sq. in. What is the 
l 

— for the column? Calculate the maximum unit-load the column would be 
r 

expected to resist. (# = 25,000,000 lb. per sq. in.) 

I ' P 

Ans. — = 310; — = 2560 lb. per sq. in. 
r a 

176. Design by Euler’s formula, a square pivot-ended timber column 20 
ft. long to resist an axial load of 60,000 lb. Use a working load equal to £ of 
the given load; assume # = 1,500,000 lb. per sq. in. 

177. What is the ratio of the strength of an ideal slender solid cast-iron 

column 6 in. in diameter to the strength of a slender hollow cast-iron column 
having a wall thickness of 1 in., the areas of cross-section and the lengths of 
the two columns being equal? Ans. 9 to 41. 




METHODS OF OBTAINING FORMULAS 


217 


97. Methods of Obtaining Formulas for Columns Having 
Intermediate Slenderness Ratios.— The results of many tests 
of columns show that the formulas represented in Fig. 184 are 
substantially correct for small and large values, respectively, of 

- but for intermediate values of ^ (corresponding to the part of 

p 

the curve between E and F say) the test results for - fall below the 

p 

curve, the cause of the lower values of — being due to the con¬ 
ditions stated in Art. 92. Now the columns used in engineering 

structures and machines have these intermediate values of and 

r 

the conditions stated in Art. 92 have, in general, a relatively 
greater influence on the strength of columns with intermediate 
slenderness ratios than on columns with either small or large 

values of Further, as noted in Art. 91, it is practically impos¬ 
sible to obtain a rational column formula that gives quantitative 
expression to the influences of these conditions, and hence 
formulas for columns of intermediate slenderness ratios are 
always either partly or entirely empirical. The main methods 
that have been used for developing such formulas are as 
follows: 

1. To assume that the conditions stated under (2), (3) and (4) 
of Art. 92 will cause the column to bend as a whole when subjected 
to an axial load P, the maximum bending moment, at the center 
of the column, being PA, where A is the deflection of the column. 
The maximum stress in the column is considered to be the sum of 
the direct compressive stress and the bending stress due to the 
moment PA. The value of the deflection, being unknown, is 
eliminated from the equation by assuming that the relation 
between the stress and deflection in the column follows the same 
law as does stress and deflection in a beam. The widely used 
Gordon-Rankine formula, discussed in Art. 98, is obtained by this 
method; it must be regarded as essentially an empirical formula, 
for, although the statement that the stress in the column, is due 
to combined direct compression and bending is essentially correct, 
the assumption made as to the relation between stress and deflec¬ 
tion is untenable; and further, the quantitative measure of the 


218 


COMPRESSION MEMBERS, COLUMNS 


influence of the conditions stated in Art. 92 on the stress in the 
column is determined from experimental results. 

2. To determine from test data an equation which respresents 
the average of test results without any attempt at a rational ana¬ 
lysis of column action. This method gives purely an empirical 
formula; the widely used straight-line formula discussed in Art. 
99 is obtained in this way. 

3. To assume that the combined effects of the conditions stated 
in Art. 92 are equivalent, in producing stress, to a eccentricity, e, 
of loading, called an equivalent eccentricity. The bending moment, 
then, after the column bends is P(e+A) instead of PA as used in 
Euler’s equation. An expression for P is then found, from the 
elastic curve equation, which contains the maximum unit-stress 
in the column and the equivalent eccentricity, e; a value for e being 
chosen so that the formula is made to agree with experimental 
results. This method leads to a modified Euler formula called 
the secant formula which is discussed in Art. 105. 

98. Gordon-Rankine Formula. —As already noted, an ideal 
column subjected to an axial load would not bend until the Euler 
load is reached, but the deviations from these ideal conditions may 
be assumed to be equivalent, in pro¬ 
ducing stress, to an initial crooked¬ 
ness or bend in the column as a whole, 
the total deflection at the center after 
an axial load P is applied being A (Fig. 
185a). The unit-stress, then, in the 
column is due to direct compression 
and bending. 

The maximum value of the unit- 
stress is found as follows: The maxi¬ 
mum stress occurs on section AA 
(Fig. 185a)' and the stresses on this 
section must hold the external force 
P in equilibrium. Now, by introduc¬ 
ing two equal and opposite forces, 
Pi and P 2 , each equal to P, the force 
P may be resolved into a force Pi 
(Fig. 1856) acting through the cent¬ 
roid of the central section AA, and a bending couple PA. If the 
force Pi were acting alone the unit-stress s 1 developed on the area 



Fig. 185.—Stress in Column. 
















GORDON-RANKINE FORMULA 


219 


a of the section AA would be uniformly distributed over the area, 
and hence 

D P 

as\ = P or s i=—. 

a 


And if the bending couple PA were acting alone, a resisting moment 

— would be developed, the unit-stress varying as shown in Fig. 

185(6); this resisting moment would hold the external moment 
PA in equilibrium, and hence 



PAc 

or s 2 =-j. 


The actual unit-stress, then, at any point of the area, would be 
the algebraic sum of the unit-stresses due to Pi and to PA. Hence 
the maximum unit-stress s is 


S = Si+S 2 


P PAc 
a I ’ 


(146) 


and since I = ar 2 where r is the radius of gyration, the above equa¬ 
tion may be written 


s 



(147) 


Now A is eliminated from eq. (147) by assuming that it bears 
the same relation to the stress in the column as exists between 
the deflection and stress in beams. 7 This relation for beams is 
si 2 ... 

A = k— in which k is a constant depending on the kind of material 


7 This assumption does much to destroy the rational basis of the formula 
since in a beam the stress is due to bending alone, and hence even if the column 


were an ideal one. only that part of the stress 



which is due to the - 


bending moment PA should be considered to be proportional to — instead of 

c 

the total unit-stress s. Further, in order to include the effects of deviations 
from ideal conditions part of the value of A should be considered to be due to 
initial crookedness. 



220 COMPRESSION MEMBERS, COLUMNS 


of which the beams are made, the type of loading, and the end- 

l 2 

conditions. And this may be written A = </>— where <f> is a con¬ 
stant similar to k, provided that the beams are subjected to the 
same maximum unit-stress and that this stress does not exceed 
the proportional limit of the material. Therefore A in eq. (147) is 

l 2 . . . 

replaced by <£—; the resulting equation is 



(148) 


This equation is called the Gordon-Rankine column formula; 
Gordon used the least lateral dimension of the cross-section and 
Rankine introduced the radius of gyration. 

A value of <£ for columns of a given material and type of end 
connection is found from results of tests 8 of such columns, the 
value of 4> being selected so that the value of P in the equation 
agrees as well as may be with the average of the experimental 
values of the loads that caused the columns to fail. 

Values of </> thus found from experimental data vary widely; 
the values given in Table 3 are average values that have been 
used in engineering practice. The value for 4> recommended by 
Rankine for wrought-iron pin-ended columns (structural steel was 
not then manufactured) w T as -jewo an d this value is still frequently 
used in specifications for structural steel columns, as for example, 
in the Building Ordnances of many cities. The straight-line 
formula (Art. 99), however, is now used extensively in the design 
of structures and also, but to a less extent, in the design of machines. 

To determine the maximum load P that a column can resist, 
the value of s in equation (148) is made equal to the ultimate com¬ 
pressive strength of the material (for ductile material the yield- 
point should be regarded as the ultimate compressive strength), 

8 In such tests the columns are subjected to approximately the same max¬ 
imum unit-stress at failure but the value of this stress is greater than the pro¬ 
portional limit of the material, although for ductile material, such as structural 
steel, the stress in the material when the column fails probably does not exceed 
the yield-point and hence is not much greater than the proportional limit. 
However, the method of obtaining <j> } and the assumption discussed in foot¬ 
note 7 render the formula empirical rather than rational. 




GORDON-RANKINE FORMULA 


221 


TABLE 3 

Values * of < f > in Rankine’s Formula 

(The values given for “ Both Ends pivoted ” should be used for all three 
types of end condi cions except for columns that approach closely to an ideal 
column (as may be the case with columns having solid symmetrical sections) 
and for slender columns; see note below.) 


Material 

Both Ends 
Pivoted 

One End 
Fixed Other 
End Pivoted 

Both ends 
Fixed 

Timber. 

4 

2 

i 

3 0 0 0 

3000 

3 0 0 0 

Cast Iron. 

4 

5000 

2 

5000 

1 

5 0 0 0 

Wrought Iron. 

4 

2 

1 

3 6 0 0 0 

36000 

3 6 0 0 0 

Structural Steel.. 

4 

2 

1 

2 5 0 0 0 

2 5 0 0 0 

2 5 0 0 0 


* It will be noted that the values of <t> in this table are in the ratios of 4 : 2 : 1. These 
ratios were originally selected so as to make them agree with the effect of end conditions 
in very slender columns as found from Euler’s formula (see Art. 100). The columns used in 
most structures and machines, however, are not slender and in such columns experiments 
and experience have shown that deviations from ideal conditions such as crookedness in 
the column as a whole, local kinks, initial stresses, variation in the properties of the material, 
etc., are the factors that usually control the strength of the column; and that the end-con¬ 
ditions frequently have little effect, .'except in the case of slender columns and, possibly, 
columns having solid symmetrical sections. The value of <£ for columns with intermediate 
values of l/r as used in most structures and machines, should, therefore, be approximately 
the same for all three conditions of ends (see also Art. 101), and equal to, or slightly less 
than, that given for “ both ends pivoted.” 


and to determine a working load the value of s is a working com¬ 
pressive unit-stress for the material. A working value for s 
ranging from 12,500 to 16,250 lb. per sq. in. is frequently specified 
for structural steel columns (see footnote 17 and Art. 139) depend¬ 
ing somewhat on the value used for 0. For example, the Cambria 
Steel Handbook uses the formula 


P 

a 


12,500 


1 


( l -Y 


36,000 VC 


and the Philadelphia Building Laws specify 


P 

a 


H 


16,250 

1 / A 2 * 


(148a) 


(1485) 




























222 


COMPRESSION MEMBERS, COLUMNS 


Ritter’s Rational Constant .—Ritter proposed that a value of <f> 
be derived from the constants of the material instead of from test 

p 

data. Thus, since the value of — from Euler’s equation should 

a 

be equal, when ^ is large, to that from Rankine’s equation, we have, 


P 7 r 2 E 



Q 



(149) 


the value of s e being the proportional limit of the material since a 
slender column, due to the large deflection, is on the point of fail¬ 
ing when, or even before, the proportional limit is reached (see 

Fig. 182). Now since unity is small compared with ^ may 

be neglected and hence 


4 > = 


7 r 2 E’ 


(150) 


which is Ritter’s rational constant for use in Rankine’s formula for 
pivot-ended columns. This value of </> has been used to some 
extent in machine design. 


ILLUSTRATIVE PROBLEM 

Problem 178. The steel parallel rod AB (Fig. 186) of a locomotive has a 
rectangular cross-section 1 in. by 3 in. and is 6 ft. long. When the locomotive 
is starting the rod acts mainly as an axially loaded column the cross-bending 
due to the weight and to the centrifugal forces being negligible. The rod may 
be considered to be a pin-ended column as regards bending in the vertical plane 
and a fixed-ended column as regards bending in a horizontal plane; due to the 
shape of cross-section, the bar tends to bend in the horizontal plane, but the 
restraint of the pins tends to cause it to bend in a vertical plane as a pin-ended 
column. Find the working load to which the rod should be subjected, using 
a working stress equal to one-fourth of the yield-point. 

Solution .—The yield-point will be assumed to be 40,000 lb. per sq. in. and 
hence the working value of s is 10,000 lb. per sq. in. 

(a) Treating the rod as a pin-ended column: 

r2 = “ =T T7^=Tjrf 2 =A( 3 ) 2 = °- 7 5 in- 2 , 

a Oct 





PROBLEMS 


223 


whence, 



2 _ (72) 2 

~~ 0.75 


= 6910 


and 


l 

- = 83.2, 
r 



3X10,000 


1 + 


25,000 


(83.2) 


30,000 

1+1.106 


= 14,250 lb., working load. 



Fig. 186.—Parallel-rod in compression. 


(6) Treating the rod as a fixed-ended column: 


whence, 



r db 3 


db 


T2 


b 3 = T^in . 2 , 



(72)2 


iq 
1 2 


= 62,300 


and 


l 

-=249, 

r 


P=- 


30,000 


1+. 


1 


25,000 


(249) 


30,000 

1+2.48 


= 8630 lb. working load, 

and hence the maximum working load the column should resist is 8630 lb. 


PROBLEMS 

179. A low-carbon steel bar 8 ft. long and 2.4 in. in diameter is to be tested 
as a pivot-ended column, the load being applied axially. The yield-point of 
the material is found to be 36,000 lb. per sq. in. Calculate the slenderness 
ratio and the maximum load the column would be expected to resist. 

I 

Ans. - = 160, P = 32,000 lb. 
r 

180. Two 8-in. 25.5-lb. steel I-beams are latticed together, the distance 
between them being such that the moments of inertia of the cross-section 
about the two axes of symmetry are equal. The column is pin-ended and is 




















224 


COMPRESSION MEMBERS, COLUMNS 


20 ft. long. Calculate the value of-and the maximum unit-stress developed 

r 

by an axial load of 40 tons. 

Ans. - = 79.5; s = 10,700 lb. per sq. in. 
r 


181. What should be the spacing of 4X4 in. timber posts 6 ft. long, that 
support a horizontal platform which is loaded with a uniform load of 200 lb. 
per sq. ft.? Assume the platform to have no lateral support and hence assume 
the columns to be fixed at the lower end and free at the upper end; the value 
of <f> for such a column will be assumed to be four times that for a pivot-ended 
column (see Art. 100). Assume a working stress of 1000 lb. per sq. in. 


182. The steel connecting-rod on a certain engine is 3 in. by lj in. in 
cross-section and 2 ft. long. It may be assumed to be a pin-ended column 
for bending about an axis perpendicular to the 3 in. side and a fixed-ended 
column for bending about an axis parallel to the 3-in. side. What is the 
greatest pressure that may be applied on the 20-in. piston without exceeding a 
unit-stress in the rod of 16,000 lb. per sq. in.? Assume the connecting rod 
to be in its horizontal position. Ans. 170 lb. per sq. in. 



183. The member BC of the pin-con¬ 
nected truss shown in Fig. 187 is made of 
two 6-in. 15.5-lb. channels latticed to¬ 
gether; the moments of inertia about the 
two principal axes are equal. Calculate 
the maximum unit-stress in the member. 

Ans. s = 7470 lb. per sq. in. 


Fig. 187.—Truss containing 

compression members. 184 ‘ Flnd the maximum unit-stress in 

member AC of Fig. 187, assuming it to 
have the same cross-section as that of member BC described in the pre¬ 
ceding problem. 


185. A round timber column is 12 in. in diameter and 20 ft. long. If the 
column has fixed ends, what working axial load may be applied to the column 
assuming a working stress of 800 lb. per sq. in. 

186. A hollow cast-iron column has an outside diameter of 8 in., and inside 
diameter of 6 in. and a length of 15 ft. If the ends are fixed what is the max¬ 
imum unit-stress in the column when it carries an axial load of 80,000 lb.? 

Note. —The problems after Art. 99 may be solved by Rankine’s formula 
if additional problems are desired. 


99. Straight-line Formula—If a series of columns, all made of 
the same material, with the same general shape of cross-section 
and the same conditions of ends but with different slenderness 
ratios, are tested to failure by subjecting them to the same type of 






STRAIGHT-LINE FORMULA 


225 


loading, the test results will show that, for columns having inter¬ 
mediate slenderness ratios , the load per unit area (^j decreases 

approximately in the same proportion that the slenderness ratio 

P 

increases. In other words the relation between — causing failure 
and ^ is represented approximately by a straight line. 



Fig. 188.—Graphical representation of straight-line column formula. 

For example, Fig. 188 shows the experimental results obtained 
from tests 9 of pin-ended axially-loaded structural-steel columns 
having H-sections. The ordinates represent loads causing failure 

divided by the cross-sectional area (^j and the abscissas repre¬ 


sent the slenderness ratios of the columns 


The variation 10 in 


the results for the three columns in each group indicates the 
effects of the factors (accidental eccentricities, initial crookedness, 
initial stresses, etc.) discussed in Art. 92. As shown in Fig. 188, 

for values of - between 50 and 150 approximately, a sloping 
r 

straight line fits the test results fairly well; for values of - less than 

9 Made at the Watertown Arsenal; see “ Tests of Metals,” 1909. 

10 The variations in some tests are much greater than those shown in Fig. 
188, particularly in test of large built-up sections. 

















































226 


COMPRESSION MEMBERS, COLUMNS 


about 50 the load per unit-area causing failure is considered to be 
constant and equal approximately to the yield-point of the mate¬ 
rial, even though laboratory tests of columns having solid section 

P 

sometimes show (as in Fig. 188) that the actual — causing failure 
is greater than the yield-point of the material; tests of built-up 

p 

columns, however, rarely give values of — greater than the yield- 
point. 

P P t r 2 E 

Further, the values of — found from Euler’s formula, — = 

’ a a /7N 


r 


for ideal pivot-ended column are less 11 than the values of — found 

• a 

from tests, when ^ is relatively large (170 to 200). T. H. Johnson 12 

found from a careful study of the results of tests of slender columns 
that if 7 r 2 in Euler’s equation were replaced by 16 the resulting 
P 1QE 

equation, — = - 77 V 2 ? wou ld agree well with the test results, for pin- 


* © 


ended slender columns. The curve representing this equation is 
showm in Fig. 188. Thus, the straight line AB, in Fig. 188 repre¬ 
sents the test results only between certain limiting values of K 
Now the equation of a straight line (similar to AB, Fig. 188) is 

y = Cx+s y , 

in which C is the slope of the line, being negative in Fig. 188, and 
s y is the intercept on the 7/-axis. Therefore the equation of the 
line AB (Fig. 188) is 

- = Sy~C '.(151) 


11 This is probably due to the fact that a pin-ended column does not act 
as an ideal pivot-ended column; the friction of the pin exerts an end-moment 
which causes the column to act with an end restraint intermediate between a 
pivot-ended and a fixed-ended column; the influence of this end restraint 
being relatively large in the relatively slender columns. 

12 See Transactions of American Society of Civil Engineers, Vol. 15, 1886, 
p. 517. 



STRAIGHT-LINE FORMULA 


227 


which is called the straight-line column formula; it was first pro¬ 
posed by Mr. J. H. Johnson 13 in 1886. 

P 

But as already noted, this equation never reduces to — = s y since 

the value of ^ in the equation cannot be less than that correspond¬ 
ing to the point A (Fig. 188). 

The above equation for the straight line 14 shown in Fig. 188 is 

- = 37,500-125-,.(152) 

a r 

and this equation agrees reasonably well with most of the available 
test data 15 obtained from tests on structural-steel pin-ended, 
axially-loaded columns in which the full strengths of the col¬ 
umns were developed; that is, the columns did not fail due to 
local wrinkling or to weak end connections, etc. 

P 

Since the value of — given by equation (152) is the load per 

P 

unit-area that causes failure, a working value of — may be found 

by dividing the right-hand member of the equation by a factor 
(to make allowance for uncertainty of loads, etc., see Art. 7)— 
a so-called factor of safety. A value of 2.5 is recommended 16 for 


13 Transactions of American Society of Civil Engineers. Vol. 15, 1886, 
p. 517. 

14 Prof. J. B. Johnson found that a parabola fitted the tests results even 
better than a straight line, and the following equation of a parabola was found 
by the Column Committee of the A. R. E. A. to agree well with test results 
for structural-steel pin-ended columns 


P 5 l 

- = 32,500-- 
a 8 \r 


By applying a reduction factor of 2.5 the working or design formula becomes 

P 1/A 2 

- = 13,000-- - . 

a 4\r/ 


15 See Report of Sub-Committee on Iron and Steel Structures of the 
American Railway Engineering Association, Vol. 21, Jan., 1920. Appendix B. 

16 Since this value is based on the yield-point it is equivalent to a value 
of about 5 based on the ultimate strength; the value commonly used for 
tension members and beams is 2 based on the yield-point and hence about 4 
based on the ultimate strength. For a discussion of this matter see Art. 
139. 



228 


COMPRESSION MEMBERS, COLUMNS 


structural steel columns by a joint committee of the American 
Railway Engineering Association (A. R. E. A.) and the American 
Society of Civil Engineers (A. S. C. E.); the working load per unit- 
area then, is 

- = 15,000 - 50-, .(153) 

a r 

p 

with a specified maximum allowable value for -- of 12,500 lb. per 

sq. in. This equation is represented by the line MN in Fig. 188. 
Thus, the design formulas 17 for structural-steel pin-ended 18 
axially-loaded columns so designed that they do not fail due to 
local wrinkling or to weak end-connections are: 


P l 

— = 12,500, if - is not greater than 50, . . . (154) 

- = 15,000 — 50,- if - lies between 50 and 150 . (155) 

a r r 


P 16 E l . . 

— =- 2 n - is greater than 150. 


© 


(156) 


However, the values of — found from the last two of the above 
a 


equations differ but little for values of - between 150 and 210, as 


17 The following formulas were formerly given in the specifications of 
the A. R. E. A. and are still widely used for the design of columns in build¬ 
ings, etc. 

P l 

— = 14,000 if - is less than 50, 
a r 


P l l 

— = 16,000 — 70 - if - is between 50 and 120. 
a r r 


Thus the later specifications changes the former by reducing the maximum 
permissible unit-stress and by deducting less for the effect of length. It is 
important to note that the reduction in the permissible stress in columns has 
been accompanied by a tendency to increase the permissible stresses in beams 
and tension members, and this is consistent with the discussion in Arts. 91 
and 92. 

18 The effect of end conditions is discussed in Art. 100; but specifications 
usually make no allowance for the end restraint that occurs due to riveted 
end-connections as used in buildings, bridges, cranes, etc. 




STRAIGHT-LINE FORMULA 


229 


indicated in Fig. 188, and hence the second of the above equations 
may be considered fairly reliable for values of ^ between 50 and 210. 

In designing a column, by the straight-line formula, to resist 
a given load, the dimensions of the cross-section may easily be 
found directly from the formula, provided that the area is circular 
or square, but the dimensions of sections of columns made of 
rolled shapes it is convenient to use the method of trial and error, 
since the calculations involve quadratic equations. Steel com¬ 
panies’ hand books give safe loads on columns having various sec¬ 
tions as calculated by different formulas, and these values may be 
used to secure the first trial value of the section when employing 
a formula different from that used in the handbook. 

Timber and Cast-iron Columns .—The following straight-line for- 

P 

mulas giving working values of — are representative of those in 
common use. 

P l 

— = 1000—6-, for structural timber, . . . (157) 

a r 

- = 9000—40-, for cast iron.(158) 

a r’ 


ILLUSTRATIVE PROBLEM 


Problem 187. A latticed, pin-ended, steel column is made of two 10-in. 
15-lb. channels (Fig. 189). The distance, h, between 
backs is such that the moments of inertia about the 
x and y axes are equal. The column is 20 ft. long. 

Calculate the slenderness ratio and the working axial 
load for the column. 


Solution. The following values are found in a steel 
maker’s handbook: 

a = 2X4.46 = 8.92 sq. in., and r = 3.87 in. 

I 240 


r 3.87 


62.0, and hence equation 155 is applicable. 


P l 

- = 15,000-50- 
a r 

= (15,000—50X62) = 11,900 lb. per sq. in. 
P = 8.92X11,900 = 106,000 lb., working load. 



Fig. 189.—Lattiecd 
column. 
















230 


COMPRESSION MEMBERS, COLUMNS 


PROBLEMS 


188. A standard 12-in., 40-lb. I-beam 8 ft. long is used as a column with 

riveted end-connections that are assumed to be equivalent to pin-ends. What 
working axial load should it carry? (From Cambria handbook; I i = 245.9 in. 4 , 
U = 10.95 in. 4 , a = 11.76 in. 2 . Ans. 117,600 1b. 

189. A square timber column 16 ft. long resists a working axial load of 
120,000 lb. Find the dimensions of the cross-section. 

Ans. 13.1 in. square. 

190. A column in a certain power-house has an unsupported length of 
22 ft. and is made up of a 12 in. by \ in. plate, to which are riveted four 5 in. 
by 3^ in. by §-in. angles as shown in Fig. 190. Find the working axial load. 



191. What size of angle should be used for the member BC of the truss 
shown in Fig. 191 (see also Problem 31 and Fig. 33). 


IT 

Fig. 192. 


192. The upper chord member of a truss (similar to AB 
or BD in Fig. 191) consists of two angles, separated | in. 
by washers (Fig. 192). If the unsupported length of the 
chord is 5 ft. and the total compressive axial stress is 65,000 
lb. find the size of angles required. 

Ans. 5-in. by 3^-in. by j^-in. angles. 


Note: The problems after Art. 98 may be solved by the straight-line 
formula if additional problems are desired. 


100. Effect of End-conditions.—I. On Ideal Slender Col¬ 
umns. —The more important ideal • end-conditions that are 
approached more or less closely in the end-connections of struc¬ 
tural columns are illustrated in Fig. 193, and may be described 
as follows: 

(а) Both ends free to turn but not free to move laterally. 
Fig. 193(a); a column subjected to such end-conditions is usually 
called a pivot-, round-, or hinge-ended column. 

(б) Both ends fixed so that the tangents to the elastic curve at 
the ends are parallel to the original axis of the column, Fig. 193(6). 




















EFFECT OF END CONDITIONS 


231 


(c) One end fixed and one end free to turn but not free to move 
laterally, Fig. 193(c). 

( d ) One end fixed and one end free from all restraint, Fig. 
193(d). 

The axial load P that will cause the column shown in Fig. 
193(a) to bend (considered to be the maximum load a slender col¬ 
umn can resist, see Art. 95) is P = - J^ . 



Fig. 193. —Effect of end-conditions on ideal slender columns. 


In Fig. 193(5) the points of inflection, A and C, are at a dis¬ 
tance of from the ends, and hence the middle half, ABC, is a 
column of the same type as that in Fig. 193(a). Thus the max¬ 
imum load P for a fixed-ended slender column is 


P = 


ir 2 EI ±tt 2 EI 


© 


l 2 


(159) 


Therefore, a fixed-ended slender column of length l will carry as 

great a load as a pivot-ended column of length that is, a fixed- 

ended slender column of a given length is four 19 times as strong 
as a pivot-ended column of the same length. 


19 See footnote 22. 

































232 


COMPRESSION MEMBERS, COLUMNS 


In Fig. 193(c) the inflection point C is at a distance of approx¬ 
imately 0.7 1 from A, and the part of the column ABC is of the same 
type as that of Fig. 193(a). Hence, 

D 7 r 2 EI 2 tt 2 EI . , , 

P= (Q 7 Q 2 = 12 - approximately. . . (lbO) 


Therefore a slender column with one end fixed and the other end 
pivoted is approximately twice as strong as a column of the same 
length having both ends pivoted. 

In Fig. 193(d) the curve AB assumed by the column corre¬ 
sponds to the portion AB of Fig. 193(a) and hence Euler’s equa¬ 
tion for the column in Fig. 193(c) is 


tt 2 EI = 1 TV 2 El 
(21) 2 4 l 2 * 


(161) 


Therefore, a slender column fixed at one end and free or unre¬ 
strained at the other is only one-fourth as strong as a pivot-ended 
column of the same length. 

II. On Columns as L t sed in Structures and Machines.— 
The end connections of columns used in structures and machines 
do not as a rule permit the ideal conditions discussed above to 
occur. The more common types of structural columns are: 

Pin-ended Columns .—A column in a pin-connected structure 
is usually assumed to approximate closely to a pivot-ended 
column as regards bending in a plane perpendicular to the axis of 
the pins, and to approximate more or less closely to a fixed-ended 
column as regards bending in a plane parallel to the axis of the 
pins (see discussion under fixed-ended columns below). Tests 20 
indicate that the friction of the pin, particularly when the pin is 
relatively large and the load relatively small, causes the column 
to act as a fixed-ended column but when subjected to larger loads 
it acts as a pivot-ended column, and hence in practically all cases 
the friction should be neglected. Pin-ended columns are commonly 
used for compression members in bridges, for connecting rods of 
engines, etc. 

Columns with Ends Partially Restrained .—Columns in buildings, 
bridges, cranes, etc., are frequently fastened to other members by 


20 See tests made by James Christie, Trans. A. S. C. E., 1883, pages 85-122; 
also “ Tests of Large Bridge Columns ” by Griffith and Bragg, Technologic 
Paper No. 101, Bureau of Standards, 1918. 





RANKINE’S FORMULA FOR RESTRAINED ENDS 233 


riveted connections but columns with these kinds of end connec¬ 
tions cannot have fixed ends since the connections and also the 
supports to which the column is attached are never rigid. The 
amount of restraint depends on the relative stiffness of the column 
end-connections and of the supports to which the column is 
attached, the stiffer the connections and supports the more the 
column is restrained at its ends. 

As a rule it is impossible to determine in any specific case to 
what extent the ends are restrained, and in engineering practice the 
assumption is frequently made that all columns have pivot-ends. 
However, if the column is known to be well restrained allowance 
may be made for the increase in strength as indicated in Table 3 
and in Art. 102. 

Flat-ended or Square-ended Columns .—This type of column 
bears against approximately plane surfaces of end supports or foot¬ 
ings. If the surfaces of the ends of the column and of the footings 
were true plane surfaces and made perfect contact over the whole 
end surface of the column, flat-ended columns would approach 
closely to fixed-ended columns provided that the end supports 
were relatively rigid. However, there is always much uncertainty 
as to the way the column bears against the end supports, 
the unevenness of bearing causing, at working loads, an eccen¬ 
tricity of loading, and hence a so-called flat-ended column is, as 
a rule, considered to be little if any stronger than a pivot-ended 
column of the same slenderness ratio. Timber columns are com¬ 
monly used with so-called flat ends. 

101. Rankine’s Formula for Columns with Restrained Ends.— 
In Rankine’s formula the effect of end-conditions is introduced in 
the value of 4> as indicated in Art. 98 and in Table 3. Now <£ 
entered the equation in the term that was assumed to measure the 
resistance to bending only; further, end restraints affect the 
resistance to bending mainly, therefore, it has usually been assumed 
that the values of <f> for the various types of end restraints (see 
Table 3) are proportional to the effect of the same end restraints 
on ideal slender columns as found under I of Art. 100 above, since 
in slender columns bending is the only action considered; thus, <£ 
for fixed ends has usually been assumed to be four times that for 
pivot-ends, etc., as is shown by the values in Table 3. 

But (as already stated after Table 3) the results of tests of col¬ 
umns having intermediate values of slenderness ratios indicate 


234 


COMPRESSION MEMBERS, COLUMNS 


that the type of end conditions may have a much less effect on the 
strength of the columns than does the deviations from the assumed 
ideal conditions (see Art. 92), such as crookedness in the column 
as a whole, local kinks, initial stresses, variations in the properties 
of the material, etc. For example, some tests of steel columns 
having slenderness ratios less than 100 show that the strengths 
of the columns may be influenced but little by any of the end 
conditions commonly used. 

Therefore, except for columns that are relatively slender and 
for columns that approach closely to ideal columns as may be the 
case for members with solid symmetrical sections, it is well to 
assume that the value of 4> is the same for the different conditions 
of ends and equal to, or slightly less than, that given for pivot- 
ended columns (see Table 3 and formulas (148a) and (1486). 

102. Straight-line Formulas for Fixed-ended Columns.—As 
noted in Art. 100 the strength of an axially loaded slender ideal 
column with fixed or flat ends is four times that of a similar column 
having pivot ends. The strength of an ideal column having a 
very small slenderness ratio, on the other hand, is approximately 
the same for fixed ends as for pivoted ends since the strength of the 
column depends only on the strength of the material. Further, 
since the fixing of the ends increases the resistance of the column to 
bending chiefly, the influences of fixed ends would be expected to 

increase with the increase of and test results 21 indicate that this 

r 

is true. 

P l 

If test results for — and - for fixed-ended structural-steel col- 
a r 

umns are plotted, in a way similar to that used in Fig. 188 for pin- 
ended columns, a straight line is found to fit approximately the 
average of the test results, the equation of the line being 

^ = 37,500 - 90^.(162) 

If a working factor of 2.5 is used, as was done with the equation 
for pin-ended columns, the resulting equation giving the working 
load for structural-steel fixed-ended columns is 

- = 15,000 — 40-, .(163) 

a r v ' 

21 See Bulletin American Railway Engr. Assoc., Vol. 21, Jan., 1920, Ap¬ 
pendix B. 


STRAIGHT-LINE FORMULAS FOR FIXED-ENDED COLUMNS 235 


in which -- must not exceed 12,500 lb. per sq. in.; thus for values of 
l P 

- less than 62.5 the equation is, — = 12,500 lb: per sq. in. The above 
equations are recommended by the American Railway Engineer¬ 
ing Association. The upper limit for ^ to be used in eq. (163) is 

about 200 and for larger values than 200 Euler’s equation 22 is. 
applicable. 

Equation (163) should be used only in cases where the restraint 
of the end connections are comparable to that offered by the rela¬ 
tively rigid heads of a testing machine when making a good con¬ 
tact with the end surfaces of the column. 

Buckling of Flange of Beam .—The compression flange of an 
I-beam or built-up girder is similar to a column with ends par¬ 
tially restrained, and hence may fail by deflecting sidewise. It is 
customary, therefore, to restrict the maximum compressive unit- 
stress in the angles or angles and plates which compose the com¬ 
pression flange of a built-up girder; the maximum allowable 
unit-stress, is stated in the A. R. E. A. specifications to be equal 
p . 

to — in the formula 
a 

- = 14,000 - 200j,.(164) 

a b’ 

in which l is the distance between lateral supports and b is the 
width of the flange. This equation becomes 

-=14,000 - 58-,.(165) 

a ’ r’ 


if the flange be regarded as rectangular, since then b 2 =12r 2 and 
6 = 3.46r. 

103. Straight-line Formula for High Carbon and Alloy Steel 
Columns.—Formulas 154, 155 and 163 apply to columns made of 


22 Mr. T. H. Johnson found from a careful study (Trans. Am. Soc. C. E.,. 
Vol. 15, 1886, p. 517) of test data for slender columns having flat ends that 
approached closely to fixed ends, that Euler’s equation was approximately 


P 

a 


2 ~~ instead of 


0 


4-7T 2 E 



as found for ideal slender fixed-ended columns. 




236 


COMPRESSION MEMBERS, COLUMNS 


steel having an ultimate tensile strength of 55,000 to 65,000 lb. 
per sq. in. and a yield-point of at least one-half of the ultimate 
strength, since these are the values stated in the specifications of 
steel for most structural purposes. 

Now if steel having a higher strength than this is used, the 


strengths of columns having small values of ^ will be increased in 

about the same ratio as the yield-points (or proportional limits) 
are increased, whereas the strength of the very slender 
columns will be increased very little, if any, since the strength 
of such columns depends on the stiffness (modulus of elas¬ 
ticity) of the material (not the strength of the material) 
and the stiffness of all grades of steel is practically con¬ 
stant. 23 

There are at present not sufficient test data from which to 
determine the constants in a straight-line formula for columns of 
but the few test results 24 available indicate 
that high-strength steel is of little 
advantage for slender columns and 
that, in accordance with the above 
reasoning, the formula for struc¬ 
tural steel columns should be 
modified for use with high- 
strength steel by increasing the 
constant C more than the constant 


high-strength steel 



s y , otherwise the value of — will be 


Fig. 194. 


too high for the larger values of 
In other words, if the line AB (Fig. 194) represents the straight- 


line formula for ordinary structural steel the line CD would repre¬ 
sent the formula for high-strength steel. • 

For example the safe load for a column made of high-strength 


23 Tubular steel struts sometimes used on airplanes have large values of - 

r 

and hence the struts will be little if any stronger when made of special alloy 
heat-treated steel than if made of ordinary low-carbon steel, although most of 
the other steel members of the airplane will be much stronger if made of the 
high-strength steel. 

24 Bull. Am. Ry. Engr. Assoc., Vol. 21, Jan., 1920, App. B. 






ECCENTRICALLY LOADED COLUMNS 


237 


steel (having a proportional limit of about 60,000 lb. per sq. in.) 
as given by the following straight line equation 


= 22,500 


90-, 

r 


(166) 


with a maximum value for — of 18,750 lb. per sq. in., is comparable 


with the value of P in the equation — 


15,000 — 50 - (with a max¬ 


imum of 12,500 lb. per sq. in.) for a column made of structural 
steel, as given in Art. 99. 

104. Eccentrically Loaded Columns.—In the preceding articles 
the column was assumed to be axially loaded. Many columns, 
however, are subjected to eccentric loads, 
and the Rankine formula and the straight- 
line formula may be modified to take 
account, approximately, of the eccentric¬ 
ity of loading as follows (the modified 
Euler’s equation is discussed in Art. 105): 

Modified Rankings Formula .—In Fig. 

195(a) let the column be subjected to a 
load P having an eccentricity e. This 
load may be resolved into an axial load 
Pi (equal to P) and a couple Pe (Fig. 

195). Now the load Pi if acting alone 
would cause a unit-stress si on the con¬ 
cave side which, according to Rankine’s 
formula, is 

PL . .I 2 
s i=—( 
a \ 





(a) 

Fig. 195.—Eccentric load 
on column. 


and the bending couple Pe if acting alone would cause, and be held 
in equilibrium by, a resisting moment Pe = see Fig. 
1955). The unit-stress, s, on the concave side, then, is 
S = Si + S2 

Pec 


P ec\ 
r 2 r 2 /’ 


-(!+* 


(167) 










238 


COMPRESSION MEMBERS, COLUMNS 


in which r is the radius of gyration of the cross-section about the 
centroidal axis, and c is the distance from the centroidal axis to 
the fiber on which the unit-stress s occurs. 

Modified Straight-line Formula .—According to the straight- 
line formula for axial loads (Art. 99) the maximum average unit- 

stress ^0 to which the column may be subjected is 

P -=Sy--C l - .(168) 

a y r 

That is, a column with a given ^ may be considered to be an 

axially loaded compression block in which the allowable unit- 

stress is (^Sy—C-^j , provided that the value of ^is between certain 

imiting values, as explained in Art. 99. 

Now if the load P is applied with an eccentricity e, as in Fig. 
195(a), the unit-stress due to the bending moment Pe would be 

Me Pec Pec 

S2 ~T I W 2 ’ 


and the maximum unit-stress in the column would be increased by 

P 

this amount as indicated in Fig. 195(6), and hence the value of — 

must be reduced by the amount $ 2 . 

Thus, 

-=U-c-)-~ 

a \ r] ar z 

or .(169) 



That is, a column with a given - and subjected to a load P having 

an eccentricity e may be treated as an axially loaded short com¬ 
pression block for which the allowable unit-stress is 






PROBLEMS 


239 


provided that the value of - is between certain limiting values as 
explained in Art. 99. 


PROBLEMS 


193 . A timber column supports one end of a beam as indicated in Fig. 196. 
The column has a square cross-section 12 in. by 12 in. and is 12 ft. long, and is 
assumed to have ends equivalent to pin ends. If the load is assumed to act 
with an eccentricity of 4 in., what working load should the column support? 





Fig. 196. —Eccentric load on column. 


r 

1£-1 

u,,» 

4 

bz 722 

. 4 * 

7223 

1 ~ c 

s 3' ' 

i 3 

ps 

=a 

' 77?i J 

u 

r 

\' r rr\ 


12 


r 

1 Xl 

1 4 


Fig. 197. 


Details of 
Channel 
Cl= 5 sq. in 
I 3 = 80 in. 4 
I 4 = 3.0 in. 4 
X = 0.6 in 


194 . A pin-ended steel column having the cross-section shown in Fig. 197 
is subjected to a load of 40 tons having an eccentricity of 2 in.; the load acts at 
A, Fig. 197. Calculate the maximum unit-stress in the column. 

195 . A column is built up of two 10-in., 15-lb. channels laced together 
(see Fig. 189) so that the distance between backs is 6.33 in. (the moments of 
inertia about the two principal axes are then equal). The column is 18 ft. 
long. Calculate the working load if the eccentricity is (a) 0 in., (6) 2 in., 
(c) 4 in. 


105. Equivalent Eccentricity. Secant Formula. —As stated in 
Art. 97, one method of taking account of the effect of crookedness, 
lack of homogeneity of material, initial stresses, unintentional or 
accidental eccentricity, etc., is to assume that these conditions 
are equivalent, in producing stress, to a positive eccentricity of 
loading on a straight column of homogeneous material free from 
initial stresses. Thus, if the eccentricity be denoted by e, the 
bending moment at any section of the column is P(e-\-y), as shown 
in Fig. 198(a), instead of Py as was the case in the derivation of 
Euler’s column formula for an axially-loaded column (Art. 95). 
Now, as shown below, this change in the expression for the bend¬ 
ing moment leads to a modification of Euler’s formula, called 
the secant formula, from which the maximum unit-stress (and also 
maximum deflection) caused by a given load and eccentricity can 






















240 


COMPRESSION MEMBERS, COLUMNS 


be found; the stress and deflection do not occur in Euler’s equa¬ 
tion for an axially-loaded column. Further, the secant formula, 
can be used for eccentrically loaded columns also, since e then is 
the actual eccentricity plus the equivalent eccentricity. 

Secant Formula .—The secant formula may be derived as 
follows: Let Fig. 198(a) represent a column of length l subjected 
to a load P having an eccentricity e. 
Now the equation of the curve assumed 
by the column is the same as that of a 
round-ended column of length L (Fig. 
1986) subjected to an axial load P, the 
part MN of the column (Fig. 1986) 
being in the same state whether the 
load P is applied at A with eccentricity 
e or at C with no eccentricity. Now, as 
stated in Art. 95, when the load P is 
applied at C as an axial load on a pivot- 
ended column, the column deflects 
when the load reaches a value Pi (say) 
slightly greater than that of P given by 
■ 2 EI 



( 6 ) 

Equivalent 


the Euler formula P = 


l 2 


Further, 


eccentricity 


it was stated in Art. 95 that although 
there is a definite deflection corre¬ 
sponding to each value of Pi, a very small increase in the load 
causes a very large increase in the deflection, and hence we may 
assume that when the column bends, its position may be that of 
any one of the several curves CD in Fig. 198(6). But, for any 
given eccentricity e and length l, as indicated in Fig. 198(6), there 
is only one curve that the column can assume, and the maximum 
deflection A and the maximum unit-stress due to the bending 
moment P(e-fA) do not disappear from the elastic curve equation 
as was the case in the treatment of the axially-loaded slender 
column in Art. 95. 

Now, from Art. 95, the equation of any one of the curves CD 
(Fig. 1986) is 

T 


pA sm 


4k 


(170) 


if the axes are chosen with the origin at C ; but if the axes are 


















EQUIVALENT ECCENTRICITY. SECANT FORMULA 241 


chosen with the origin at the center of the column length (at O', 
Fig. 198), the equation is (see footnote for Art. 95). 


y = A cos 



(171) 


But y in this equation is the deflection of the pivot-ended column 
of length L (Fig. 1986) and hence if y is made to represent the 
deflection of the eccentrically-loaded column as shown, Fig. 198(a), 
the equation of the elastic curve assumed by the column is 

y+e= (A+e) cos .(172) 

Now y = 0 when x = ^. Therefore, 

e = (A+e) cos • • • • (173) 


and hence the maximum deflection of the eccentrically-loaded 
column is 


A=e ( sec WS -1 )’ 

A=e ( sec ^\S _1 )- 


or 


The maximum bending moment is 
M = P( A+e) 


= p|^sec 


2 r\aE 


- 1+6 


p 1 i- 
= Pe sec tta “U* 
2 rxaE 


The maximum fiber unit-stress s is 

P , Me 

S = -b-y- 

a 1 


(174) 


(175) 



Pec sec pr- 
2 r 


ar 2 




i+ se 4 



(176) 






242 


COMPRESSION MEMBERS, COLUMNS 


I P 

The expression — is an angle expressed in radians and is 

called the Eulerian angle. The above equation may also be 
written in the form 


,ec l IP' 
1+ r 2 S6 ° 2 r\aE 


(177) 


The above equations are valid only for values of s within the 
proportional limit of the material since the elastic curve equation 
and the flexure formula are used in their derivations. But within 
the proportional limit they are valid provided that the column 
acts as a unit and hence does not fail due to local wrinkling or 
weak end connections. And when these conditions are satisfied 
the equivalent eccentricity e may be selected so that the calcu- 
p 

lated values of -- for axially loaded columns are made to agree as 

well as may be with experimental results. The secant formula 
has been presented from time to time in technical literature as 
being as satisfactory a solution of column action as is likely to be 
obtained, but due largely to its relative y complicated form and to 
insufficient experimental data for determining reliable values of 
the equivalent eccentricity it is not in general use. 

When the dimensions (a, c, r, and l) of the column are given 
and a value of e is known or assumed, the load P required to cause 
a specified unit-stress, s, may be found from equation (176). And 
likewise the area a needed to keep the unit-stress from exceeding 
the prescribed value, s, when the column is subjected to a given 
load, P, may be found if values of c, r and s are known. But in 
equation (176) since the quantities P and a are not expressed 
explicitly in terms of the other quantities, the solution of the equa¬ 
tion is most conveniently made by trial and error. To aid in the 


solution of equation (176) values of 


' IZ 

\al<] 


/— 7 =» for various values of 


— and - are given in Table 4 for use in the investigation of steel 
columns (E = 30,000,000 lb. per sq. in. for steel). 


Values of e and of —The value of the equivalent eccentricity 

e to be used in equation (176) for axially loaded columns must be 



EQUIVALENT ECCENTRICITY. SECANT FORMULA 243 


assumed, the value depending on the judgment of the engineer. 
Moncrief , 25 from the analysis of the results of many tests of columns 

6 C 

found the value of to range from 0.15 to 0.6, and he recommended 

the larger value to be on the safe side. He found that this value 
applies alike to cast iron, wrought iron, mild steel, high carbon 
steel and several kinds of timber. 


TABLE 4 


Values of sec 


L IZ 

2r\ aE 


in Secant Formula 


Values of 

r 



50 

60 

70 

80 

90 

100 

110 

120 

130 

140 


5,000 

1 

.05 

1 

.08 

1.11 

1.15 

1.20 

1.25 

1 

.32 

1.40 

1.50 

1.62 


8,000 

1 

.09 

1 

.14 

1.19 

1.26 

1.35 

1.46 

1 

.60 

1.79 

2.05 

2.41 


10,000 

1 

.12 

1 

.17 

1.25 

1.34 

1.47 

1.63 

1 

.86 

2.18 

2.65 

3.46 


14,000 

1 

.17 

1 

.25 

1.37 

1.54 

1.77 

2.12 

2 

.68 

3.69 

6.02 

16.4 


18,000 

1 

.22 

1 

.35 

1.53 

1.79 

2.21 

2.95 

4 

.51 

9.86 




20,000 

1 

.25 

1 

.40 

1.62 

1.95 

2.51 

3.62 

6 

.65 

47.40 




22,000 

1 

.28 

1 

.45 

1.71 

2.13 

2.90 

4.65 

12. 

3 




a, j e 25,000 

1 

.33 

1 

.54 

1.88 

2.47 

3.73 

7.87 






«+-( 

0 

28,000 

1 

.38 

1 

.64 

2.08 

2.93 

5.15 

22.90 







30,000 

1 

.42 

1 

.72 

2.23 

3.32 

6.79 







cj 

32,000 

1 

.46 

1 

.79 

2.41 

3.83 

9.86 




Values of 

> 

35,000 

1 

.52 

1 

.92 

2.73 

4.93 

28.70 





l i 

P 


38,000 

1 

.59 

2 

07 

3.13 

6.80 





sec —\ — 

2 r\ aE 


40,000 

1 

.63 

2. 

18 

3.46 

9.07 









42,000 

1 . 

.68 

2. 

31 

3.87 

13.30 









45,000 

1 . 

76 

2. 

51 

4.68 

43.00 









48,000 

1 . 

85 

2. 

76 

5.88 










50,000 

1 . 

91 

2. 

95 

7.06 










For a circular cross-section, if -3 = 0.6, e = 0.3 r and hence 

for a column having a circular section 10 in. in diameter, e would 
equal f in. 

25 Trans. Am. Soc. Civ. Engrs., Vol. 45, 1901, p. 334. 

















244 


COMPRESSION MEMBERS, COLUMNS 


Prichard 26 proposes the following: 


ec _ J__ 1 l_ 

~~^2 T7T i 


(178) 


r* 10 700 r 

And when this is applied to ordinary structural-steel pin-ended 

c 

columns in which the value of - is usually about -J, 

e = 0.07r+0.001-.(179) 

r 

Thus a column having the relatively large ^ of 100 (r = 5 and l = 500) 

the value of e is 0.85 in. 

Basquin 27 proposes the following: 


^ = 0 . 01 + 0 . 001 - 


(180) 


In Fig. 199 are shown the curves representing the secant formula 



40 80 120 100 200 240 280 

Values of -~ 


Fig. 199.—Effect of values of — on secant column formula. 

7*2 

cc 

for steel columns, with different values of the eccentric ratio - 5 : 

r 2 

the yield-point of the steel is assumed to be 40,000 lb. per sq. in. 

26 Trans. Am. Soc. Civ. Engrs., Vol. 61, p. 173. 

27 Journal Western Society Engrs., Vol. 18, 1913, p. 457. For a compila¬ 
tion of the recommendation of many authorities see Salmon’s “ Columns,” 
Oxford Technical Publications, p. 148. This book also gives an excellent 
discussion of the factors entering in the various column formulas. 









































PROBLEMS 


245 


PROBLEMS 

196 . A column having the same cross-section as the column described 
in Prob. 172 (see Fig. 183) and a length of 25 ft. is subjected to an axial load 

of 120,000 lb. Find the maximum unit-stress, assuming that — = 0.5. 

r 2 

Ans. s = 8530 lb. per sq. in. 

197 . The column described in Prob. 183 is subjected to an axial load of 
73,000 lb. Find the maximum unit-stress in the column; use eq. (178) for 

finding a value of p. Ans. s = 9900 lb. per sq. in. 

198 . Solve Prob. 180 by the secant formula using a value of 0.6 for —. 

r 2 


199 Change the length of the column described in Prob. 192 to 6 ft. and 


solve by the secant formula using a value for — as given by eq. (180). 

r 2 


106. Built-up Steel Columns. —In deriving the column for¬ 
mulas in the preceding articles it was assumed that bending or 
flexure existed in the column as a whole, the deflection of the axis 
of the column being the result of initial eccentricity of the load, 
lack of homogeneity in the material, a general bend or lack of 
straightness in the column as a whole, or a combination of two or 
more of these conditions. Now the resistance of the column to 
bending is increased by giving the cross-sectional area as large a 

radius of gyration as possible, thereby causing ^ to be small. This 

is accomplished in built-up columns by using thin component parts 
(plates, channels, angles, etc.), and placing the parts as far from 
the neutral axis as practicable, thereby tending to produce a 
flimsy column. To what extent the cross-section may thus be 
distended without causing the column to fail locally by wrinkling 
(secondary flexure) cannot be definitely known but it is probable 
that some of the disastrous failures of columns have been due to 
wrinkling 28 or to weak end connections. 

In order to help prevent local failure it is the practice in good 
design of built-up columns that the slenderness ratio of each 


28 For a column formula which takes account of the possible failure by 
wrinkling, see “ Strength of Columns ” by W. E. Lilly, Trans. Am. Soc. Civ. 
Engrs., 1913. 


246 


COMPRESSION MEMBERS, COLUMNS 


unsupported part of the built-up column shall not exceed the - of 

the whole column. For example, the ^ of the length, p, between 

lattice bar connections of each channel in Fig. 189 should not 

exceed the - of the whole column. However, this requirement is 
r 

hardly sufficient since at failure the column is so bent that the 
maximum fiber unit-stress due to combined compression and 
bending is equal to the yield-point of the steel, and hence the com¬ 
ponent part should be designed so that it will not fail below the 

yield-point, therefore the ^ of the component parts should be lim¬ 
ited to about 40, otherwise, if a component part located near the 
point of maximum bending (near the center) is a relatively slender 

column (even though it has the same ^ as the whole column) it 

would fail without developing the yield-point strength of the 
material. 

Further, test results 29 of built-up columns show that the effects 
of local kinks, initial stresses due to riveting, etc., local eccen¬ 
tricity of loading due to the method of tying the component parts 
together (interior eccentricity) and variation in strength of com¬ 
ponent parts may, and probably have, a more important effect in 

causing local flexure or wrinkling than has the ^ of the part in 

question. Again, even though these conditions do not cause the 
column to fail by local wrinkling they may be the cause of the 
initial bending which leads to the bending of the column as a 
whole. 

It must be remembered also that high localized stress in a 
column is more serious, in general, in its effect on the strength of 
a column that it is on the resistance of a tension or flexural member. 
For these reasons the working or allowable unit-stress for built- 
up structural steel columns is low compared with that for similar 
material when used in tension members and beams. For example, 

29 “An Investigation of Built-up Columns under Load,” Bulletin No. 44 
Engineering Experiment Station, University of Illinois; also, “ Tests of Large 
Bridge Columns ” by Griffith and Bragg, Technologic Paper No. 101, Bureau 
•of Standards. 


COLUMNS SUBJECTED TO CROSS-BENDING 


247 


a maximum allowable unit-stress of 12,500 lb. per sq. in. is spe¬ 
cified for structural steel columns whereas 16,000 to 18,000 lb. 
per sq. in. is allowed for tension and flexural members, and this 

maximum value of 12,500 is reduced as ^ increases according to 

the formulas of Art. 99, for, although factors other than ^ may 

have a controlling influence, as discussed above, there is no satis¬ 
factory way to measure their effect (unless it is the equivalent 
eccentricity discussed in Art. 105 which leads to a complicated 
formula), and a column formula that is made to agree with test 
results may be used with reasonable confidence even though it may 
not be rational, provided that the column is designed so that failure 
will not occur by wrinkling or due to weak end connections, and 
that the column is similar to those used in obtaining the test 
results. 

107. Columns Subjected to Cross-Bending. —A compression 
member that is subjected to cross-bending loads may be considered 
to be (1) a beam subjected to end 
thrust as discussed in Art. 86, or 
(2) a column subjected to cross¬ 
bending loads; depending on the 
relative magnitudes of the end 
thrust and cross-bending loads, and 
on the dimensions of the member. 

In some cases both methods should 
be used in designing or investigating 
the member. 

Let Fig. 200 represent a pin- 
ended column subjected to an axial 
end load P and a cross-bending load 
Q. An approximate value of the 
unit-stress developed may be found 
as follows: 

Modified Rankings Formula .— 

According to Rankine’s formula the 

load P if acting alone wou d produce a maximum unit-stress si 
on the concave side (Fig. 2006) such that 



Fig. 200.—Column subjected to 
transverse loads. 


Si 











248 


COMPRESSION MEMBERS, COLUMNS 


Now if the deflection of the member as a beam, due to Q alone, be 

1 OP 

denoted by A (A in this case equals Art. 47) then the addi¬ 

tional stress caused by P due to the moment PA is 
PAc PAc 

S2 = - j - — -o * 

I ar 2 

This is equivalent to assuming that the column is straight and that 
the load is applied with an eccentricity of A. Further, if the 
moment due to the cross-bending load alone be denoted by M 
(in this case M = \Ql , Prob. 87) the unit-stress S 3 caused by this 
moment is 

Me Me 

S3 T o* 

I ar 2 


And if the proportional limit is not exceeded, the unit stress, s, on 
the concave side is 

S = Si + S2 + S3 



PAc Me 
ar 2 ar 2 - 


(181) 


Modified Straight-line Formula. —By use of the straight-line 
formula A may be considered an eccentricity, and equation (169) 
of Art. 104 may be used, from which the unit-stress due to the 
cross-bending must be subtracted. Hence 


P—l _M\_PAc_Mc 

a \ v r) ar 2 ar 2 ’ 


(182) 


Modified Secant Formula. —Similarly the secant formula 
(equation (169) of Art. 105) may be used, the value of e being 
made equal to A plus the assumed equivalent eccentricity e\, and 
the unit-stress due to the cross-bending load being added to the 
unit-stress due to the end load. Thus, 






(183) 


The values given by these equations err on the side of danger 
since the deflection of the column is greater than that assumed, 
but a more refined analysis, as a rule, is not practicable. 


PROBLEM 

200 . Consider the member BC in Fig. 161 to be a pin-ended column sub¬ 
jected to cross-bending and calculate the maximum unit-stress in the member. 


CHAPTER XII 


COMBINED NORMAL AND SHEARING STRESSES 

(Chapter III should be reviewed before this chapter is studied. 

108. Introduction. —The problem discussed in this chapter 
may be stated as follows: Given, at a point in a body, a shearing 
unit-stress on each of two planes at right angles to each other, and 
a normal unit-stress on one of the two planes; find the values of 
the resulting maximum normal and shearing unit-stresses at the 
same point, in terms of the given stresses, and the directions of the 
planes on which these maximum stresses occur. 

Many structural and machine members are subjected to loads 
that cause the “ given ” stresses stated above, and these stresses 
may be found from the loads, by use of the equations developed 
in the preceding chapters. The given stresses may arise in the 
following ways: 

(a) The shearing unit-stress, s s , and the normal unit-stress, s, 
may be due to central loads (Art. 3), as in a bolt loaded as shown 
in Fig. 201, in which case and s are found from the loads as 
follows: 



(184) 


and s=—. 
a 


(i b ) s s may be due to torsional loads and s to an axial load 
(Fig. 202), in which case (see Art. 26). 


Tc 


(185) 


and s =—. 
a 


(c) s a may be due to torsional loads and s to bending loads, as 
in an engine crank shaft, etc., in which case (see Art. 26 and 34) 



Tc 1 Me 


(186) 


249 





250 


COMBINED NORMAL AND SHEARING STRESSES 


(i d ) Both s s and s may occur in a beam in which case (see Art. 
40) 

s a =Y t ^y^ a anc * • • • ( 187 ) 

Further, as shown in Art. 17, if a shearing unit-stress of intensity 
s 3 occurs in one plane a shearing unit-stress equal to s s must 
occur on a plane at right angles to the first. 

109. Maximum Normal and Shearing Stresses. —Fig. 201 
shows a bolt loaded, as suggested under (a) above, so as to produce 
the combination of stresses under discussion, the stresses being 



Fig. 201. —Combined shearing 
tensile stresses. 



Fig. 202.—Combined shearing and 
and tensile stresses. 


shown on the elementary block in the bolt, and Fig. 202 shows a 
similar block subjected to the same combination of stresses by the 
method of loading suggested under (6) above. Fig. 203(a) shows 
one of these elementary blocks, enlarged, and also the forces that 



A Cs s 



X 


Fig. 203. —Combined shearing and tensile stresses. 


act on its four faces and that hold it in equilibrium; the dimen¬ 
sion perpendicular to the plane of the paper is assumed for con¬ 
venience to be unity. 

The problem is to find the value of the maximum normal and 
shearing unit-stresses resulting from this combination of stresses, 

i 7 

































MAXIMUM NORMAL AND SHEARING STRESSES 251 


and the directions of the planes on which these maximum unit- 
stresses occur. 

To Find the Maximum Normal Stress. —Let a plane be passed 
through the elementary block of Fig. 203(a) at an angle 0 with 
the plane on which the normal stress occurs. The normal and 
shearing unit-stresses on this inclined plane will be denoted by 
s'\ and s' 8 , respectively. The forces, then, that hold the block 
ABC in equilibrium are as indicated in Fig. 203(6). And since 
the sum of the components of the forces in the ^-direction and in 
.the ^/-direction must each be equal to zero, we may write, 

BCs't cos 6+BCs's sin 6—ABs s —ACs t = 0, . (188) 

and 

BCs't sin d—BCs's cos 6—ACs s = 0 . . . (189) 


By dividing each of these equations by BC and noting that 
AB . „ , AC 

~~FT 7 y = sm 0 and = cos 0 the resulting equations are: 


BC 


BC 


from (188), 
from (189) 


(s't St) cos 0-j-(s' s —s s ) sin 0 = 0, . . . (190) 

s' t sin 0— (s 3 -j-s' s ) cos 0 = 0.(191) 


But, the normal unit-stress, s' t , has its maximum or minimum 
value on the plane on which the shearing unit-stress, s’ s , is zero (see 
Art. 110 for proof). 1 Therefore, if s' s in the above equations is 
made equal to zero, and if 0' is used to denote the value of 0 when 
s' t is a maximum or minimum, the above equations become, 

(s' t —s t ) cos 0' — s s sin 0' = O, .... (192) 


s s cos 0' — s\ sin 0' = 0.(193) 


By eliminating 0' from these two equations, the maximum and 
minimum values of s' t may be found in terms of s t and s s which in 
turn may be found from the external loads by the methods already 
discussed. And by eliminating s' t from the two equations the 

1 This theorem was involved in a restricted form in Art. 35 where it was 
proved that the maximum fiber (normal) stress occurs on the section for which 
the shear is zero. It is also illustrated in Art. 16 where s n is shown to be always 
less than s, the stress on the plane where there is zero shearing stress. 



252 COMBINED NORMAL AND SHEARING STRESSES 


values of O' may be found which give the directions of the planes 
on which the maximum and minimum normal stresses occur. 

By dividing (192) by (193) the following equation is obtained: 

S t &t _ s 8 

Ss s't 
or 

s't 2 — s t s't—s s 2 = 0. 

Therefore, 


s , , = |s,± A J(| 5 ) +s, 2 , . . . 

. . . (194) 

s'f=Js < ±|V8 l 2 H-4s, 2 . . . . 

. . . (195) 


Thus, there are two principal 2 unit-stresses, one being the 
maximum normal unit-stress (in this case a tensile stress) given by 
the expression 

max. s' t = is t -\-%V s* 2 +4s s 2 , .(196) 

and the other the minimum normal unit-stress given by the 
expression 

min. s' t = is t —iVs t +4s 8 2 .(197) 

But the last term of the right side of this equation is always 
greater than and hence s't is a negative tensile stress, that is, 
a compressive stress, and may be written 

— s't=s' c = — i$*+iv / $f 2 +4s s 2 . . . . (198) 

Further, the planes on which the maximum and minimum normal 
unit-stresses occur are at right angles to each other as is proved 
below. 

If now the normal stress s t in Fig. 203(a) had been a com¬ 
pressive stress, s C} then the maximum normal stress would have 
been a compressive stress given by the equation 

max. s' c = Js c +jVs c 2 +4s s 2 , .(199) 

and the minimum normal stress would have been a negative com¬ 
pressive stress, that is, a tensile stress; namely, 

— s'c — s't = — §s c +^Vs c 2 +4s s 2 . . . . (200) 

2 A principal stress (see Art. 19) is one that occurs on a plane on which no 
shearing stress exists, and is always the maximum or minimum normal stress 
at the point under consideration. 












MAXIMUM NORMAL AND SHEARING STRESSES 253 


Direction of Planes on Which Principal Stresses Occur .—The 
value of O' in equations (192) and (193) may be found by elimi¬ 
nating s' t from these equations: 

From 193, 

, cos 0' 
sin B 

Substituting this value of s' t in (192) we have 


or, 

whence, 

Therefore, 


cos 2 0 n , . 

s s —— -Tf— s t cos 6 =s s sin 6 , 
sm 0 

s s (cos 2 6— sin 2 S) =s t sin g' cos S', 
s s cos 2 d' = \s t sin 2 O'. 

9 o 

tan 2d' =—.(201) 

s t 


Now there are always two angles, between 0° and 360°, for which 
the tangents are equal, the two angles differing by 180°. Thus, 
there are two values of 26' that differ by 180°, and hence there 
are two values of O' that differ by 90°, as 
indicated in Fig. 204. Therefore, the prin¬ 
cipal stresses occur on planes that are perpen¬ 
dicular to each other, and the direction of the 
principal planes may be found by solving 
equation (201) for O', 0' being measured from 
the plane on which the normal stress, s t , 
occurs. The position of an elementary 
block on which the principal stresses act, 
as they occur in the bolt of Fig. 201, is 
shown in Fig. 205(a). 

The Maximum Value of the Shearing Unit-stress s ' s .—As shown 
in Art. 20, the maximum shearing unit-stress resulting from two 
principal stresses is one-half the algebraic difference of the princi¬ 
pal stresses. Therefore, 

max. s' s = i(max. s' t — min. s' t ) = i(s' t ~hs' c ) 

= i(is t +iVs, 2 +4s, 2 )+(-i Sl +iV Sl 2+4 s ,2). 



Fig. 204.—Planes on 
which principal 
stresses act. 















254 COMBINED NORMAL AND SHEARING STRESSES 


Hence, 

or 


max. s' s = |Vs, 2 -f-4s a 2 , 


max. 



(202) 


And, as shown in Art. 20, this maximum shearing' unit-stress 
occurs on each of two planes that make angles of 45° with each of 
the planes on which the principal stresses occur. That is, the 




Fig. 205.—Principal stresses and maximum shearing stresses in bolt. 


planes of maximum shear bisect the angles between the planes on which 
the principal stresses occur. The planes on which the maximum 
shearing stresses occur are shown in Fig. 205(5) in which d' s = 
6 } -\- 45°. There are normal stresses also on the planes on which 
the maximum shearing stresses occur but these are not shown in 
Fig. 205(5). 



Fig. 206.—Directions of principal planes shown by areas of rupture of brittle 

material. 


The student should draw diagrams similar to Fig. 205 in con¬ 
nection with a shaft when loaded as indicated in Fig. 202, and 
explain why the cast-iron specimens shown in Fig. 206 failed as 
indicated. (See also Fig. 45). Since cast iron is weak in tension 
the failure in each case is by tension. 






























ILLUSTRATIVE PROBLEM 


255 


ILLUSTRATIVE PROBLEM 

^ Problem 201. If in Fig. 207 the axial tensile load, P, is equal to 45,000 lb., 
the twisting moment, Qq, is equal to 30,000 lb .-in. and the diameter, d, is 
equal to 3 in., find the maximum normal and shearing unit-stresses and the 
directions^ the planes on which these stresses occur. 




Fig. 207. —Stresses in bar subjected to combined tensile and torsional loads. 


Solution .—The unit-stress, s t , caused by the load P is 


P 45,000 
a ?r(3) 2 
~4~ 


= 6360 lb. 


per sq. in. 


and the shearing unit-stress, s s , caused by the torsional moment T or Qq is 


Tc 


s ‘ = 7 


30,000X1.5 

7t(3) 4 

32 


= 5660 lb. per sq. in. 


The planes on which these stresses occur at an outer fiber of the shaft, are 
shown in Fig. 207(a). 

The principal stresses, s't and s' c , and the maximum shearing unit-stresses, 
5 ' s , resulting from the above stresses are, 

max. s'*=+§ 2 +4s« 2 = | * 6360+1(6360) 2 +4(5660) 2 

= 3180+6480 = 9660 lb. per sq. in., tensile stress. 

min. s't = %s t — \ Vsp+4s s 2 = —3300 lb. per sq. in., compressive stress, 

s' a = Vsf 2 +4s s 2 = 6480 lb. per sq. in., 


and the planes on which the principal stresses occur make the angles 6' with 
the plane on which the stress s t occurs, the values of 0' being found as follows: 


tan 



2X5660 

6360 


= 1.78. 


















256 COMBINED NORMAL AND SHEARING STRESSES 


Hence 


20' = 60° 40' or 240° 40' 


and 


0 '= 30° 20' or 120° 20'. 


as indicated in Fig. 207(6). Further, the angles 0 ', that the planes of maxi- 
imum shear make with the plane on which s t occurs are 

0 ' s = 0 '+45° 

= 75° 20' and 165° 20'. 

These planes are shown in Fig. 207(c). 

Problem 202. A pressure P of 10,000 lb. on the crank pin of the steel crank¬ 
shaft shown in Fig. 208(a) is required to turn the shaft at constant speed when 
the shaft is subjected to a constant resisting torque. If the diameter, d, of 
the shaft is 4 in., find the maximum combined normal and shearing unit- 
stresses at the section AB. Also find the ratios of the maximum normal and 
shearing stresses to the corresponding yield points of the material if the tensile 
yield point is 40,000 lb. per sq. in. P acts perpendicular to the pin and to 
the crank. 



Fig. 208.—Stresses in crankshaft. 


Solution .—By considering a free body diagram of the part of the shaft to 
the right of section AB, it is evident that the only external force acting on this 
part is P , and hence the stresses at section AB are caused by P and must be 
such as to hold P in equilibrium. Now by introducing two equal, opposite, 
and colinear forces P 2 and P 3 at H (Fig. 2086), P is resolved into a bending 
load P 3 and a torsional couple P, P 2 ; and these three forces cause the same 
stresses on the section AB as the original single force P. Thus the shaft to 
the right of the section AB is subjected to combined bending and torsional 


loads. 


The bending moment at section AB due to the load P 3 is 8 P, and is held 

si 

in equilibrium by the resisting moment — on the section AB as shown in Fig. 


c 

































PROBLEMS 


257 


208(6). 
C ) is 


Thus the compressive unit-stress s at D (or the tensile unit-stress at 
Me 8X10,000 X2 


S = T = 


7T (4)< 
64 


= 12,730 lb. per sq. in. 


The twisting moment due to the forces P and P 2 is 6P and is held in 

s s J 

equilibrium by the shearing resisting moment — on section AB as shown in 

c 

Fig. 208(6). Thus the shearing unit-stress s s at D (or at C) is 


Tc 

y 


6X10,000X2 

7t(4)4 

32 


= 4770 lb. 


per sq. in. 


The maximum combined normal unit-stress at C and at D, then, is 
V = h y s 2 +4s a 2 = i X 12,730+| V (12,730) 2 +(4770) 2 
= 6365+7935 = 14,300 lb. per sq. in. 

This is a compressive stress at D and a tensile stress at C. The maximum 
combined shearing unit-stress at C and at D is 


s ' 8 = ^V$ 2 +4s s 2 = 7935 lb. per sq. in. 

If the tensile yield-point of the steel is 40,000 lb. per sq. in. and the shearing 
yield-point is six-tenths of the tensile yield-point (as found from tests, see 
Art. 140), then the maximum tensile stress is approximately 0.36 of the ten¬ 
sile yield-point, and the maximum shearing stress is 0.33 of the shearing yield- 
point. 


PROBLEMS 

203 . If the values of P and Q in Fig. 205 (a) are 8000 lb. and 6000 lb. 
respectively, and the diameter of the bolt is f in., what are the maximum normal 
(tensile) and shearing unit-stresses developed in the bolt? If the bolt is made 
of steel having a tensile yield-point of 45,000 lb. per sq. in. and a shearing yield- 
point equal to 0.6 of the tensile yield-point, what are the ratios of the maxi¬ 
mum tensile and shearing stresses to the corresponding yield-points. 

Ans. 0.56; 0.60. 

204 . A steam turbine drives an electric generator as indicated in Fig. 209. 
The diameter, d, of the shaft is 6 in. The weight of the generator is 15 tons. 









258 COMBINED NORMAL AND SHEARING STRESSES 


The steam turbine delivers 1200 horse-power to the shaft and rotates at 800 
r.p.m. Find the maximum normal and shearing unit-stresses developed in 
the shaft. 




205 . Two pulleys, A and B, are mounted on a shaft as shown in Fig. 210 . 
The driving pulley, A, transmits 10 h.p. to the shaft and the driven pulley B. 
It turns the shaft at 100 r.p.m. The smaller belt tension, T h is 320 lb. and 
the diameter, d, of the shaft is 2 in. and the diameter of each pulley is 2 ft. 
Find the maximum normal and shearing unit-stresses in the shaft. 

Ans. s'* = 15,800 lb. per sq. in.; s' s = 8400 lb. per sq. in. 

206 . The crank-pin pressure, P (Fig. 2115), perpendicular to the crank is 
5000 lb. when the shaft is turning against a constant resisting moment Qq. 
The diameter of the shaft at A, B, and C, is 3 in. Find the reactions R i and 
Rt of the bearings (which are assumed to be pivot bearings) and also the max¬ 
imum normal and shearing unit-stresses at A, B, and C. 




207 . If a shaft is subjected to combined torsion and bending, and the 
shearing yield-point of the material is six-tenths of the tensile yield-point, 
prove that the ratio of the shearing unit-stress s' 8 to the shearing yield-point 
will be equal to the ratio of the tensile unit-stress s' * to the tensile yield-point 
when the ratio of the twisting moment T to the bending moment M is equal 
approximately to 0.9. 

208 . A shaft 4 in. in diameter is subjected to an axial end thrust of 12 tons, 















































PLANE OF MAXIMUM NORMAL STRESS 


259 


and also a bending moment and a twisting moment, the twisting moment being 
equal to one-half the bending moment. The maximum normal unit-stress is 
10,000 lb. per sq. in. and the shaft rotates at 100 r.p.m. What horsepower 
does the shaft transmit? 


209 . A shaft 5 in. in diameter resists a bending moment of 120 ton-in. 
and a twisting moment of 70 ton-in. Find the magnitude of the maximum 
normal and shearing unit-stresses. 

Ans. s't=s' c = 10.5 ton per sq. in.; s' s = 5.66 ton per sq. in. 

210 . In which of the two shafts described below is the greater normal 
unit-stress developed and in which is the greater shearing unit-8tress devel¬ 
oped? Calculate the maximum unit-stresses in each shaft. 

(1) A 4-in. shaft subjected to a twisting moment of 40 ton-in. and a 
bending moment of 32 ton-in. 

(2) A 2-in. shaft subjected to a twisting moment of 7.0 ton-in. and a 
bending moment of 20 ton-in. 


110. Maximum Normal Stress Occurs on Plane of Zero 
Shearing Stress. Ellipse of Stress. —In Art. 109 it was stated 
that the maximum and minimum normal stresses occur on planes 
on which the shearing stress is zero. This fact will here be proved. 

In Fig. 212 let ABC be an ele¬ 
mentary prism at a point in any 
body; that is, AB, AC, and BC, 
are three planes passing through 
a given point in the body. Let the 
dimension of the prism perpen¬ 
dicular to the paper be unity. 

Further, let the planes AB and AC 


c 4 

»2 

U V* 

IT c 




Y 


Fig. 212. —Ellipse of stress. 


be the planes on which normal stresses only occur; the unit-stresses 
on these planes are denoted by si and S 2 respectively. On the 
plane BC there will occur both a shearing unit-stress and a normal 
unit-stress; let the resultant or actual unit-stress on the face BC 
be denoted by s r , and let the coordinates of the extremity of the 
vector s r be x and y. 

Now since the forces acting on the prism are in equilibrium, 
the sum of the ^-components and of ^/-components must equal 
zero. Hence, 


(BCs r ) cos (f) = ACsi ,.(203) 

and 

(. BCs r ) sin (f> = ABs 2 , .(204) 









260 COMBINED NORMAL AND SHEARING STRESSES 


but s r cos 4> = x, s r sin <f> ■■ 


AC 

y ’ BC = C0S 


, AB . 
and = sm 


(203) and (204) may be written as follows: 

X 

x = s\ cos 9 or — = cos 9, 
s i 

and 


v 

y = S 2 sm 9 or — = sm 9, 
S2 

from which the following equation is obtained, 

7*2 a y2 

— 2 +-^o = cos 2 0+sin 2 0 = 1. . . 

s l 2 S2 2 


Hence 


(205) 


This is an equation of an ellipse having si and S 2 , the principal 
unit-stresses, as the semi-axes, and hence the locus of the extremity 
of the vector representing the unit-stress on the face BC as this 
face is assumed to turn through 360° is an ellipse. This ellipse 
is called the ellipse of stress. 

Now, as may be seen from inspection, the unit-stress s r has its 
maximum or minimum value when it is one of the semi-axes of the 
ellipse, that is, when it occurs on a plane on which there is no 
shearing stress. 

Planes of Maximum Shearing Stress .—In Art. 20 it was shown 
that the shearing unit-stress, s' 8 , on the oblique plane BC (Fig. 
212) is 

s/ «“i(®i+«2) sin 20.(206) 

The graph of this equation is given in 
Fig. 213 in polar coordinates, s' s being 
plotted as a radius vector and 9 as 
the angle. This graph shows that the 
shearing unit-stress is a maximum on 
planes making angles of 45° with the 
planes on which the principal stresses 
si and S 2 occur and that the shearing 
unit-stress is zero on the principal 
planes. 

111. Diagonal Tensile Stress in a 
Beam.—The normal unit-stress on 
any right cross-section of a beam, 



Fig. 213.—Planes of maxi¬ 
mum shearing stress bisect 
angles between principal 
planes. 





DIAGONAL TENSILE STRESS IN A BEAM 


261 


such as section A A, Fig. 214, is the greatest at the outermost 
fiber and is found from the flexure formula, 

Me 

S= T’ 

where c is the distance from the neutral axis to the most remote 
fiber and M is the bending moment at the section considered. 
And since the shearing stress at the outermost fiber is zero (Art. 
40), the stress s is the maximum normal stress at this point of the 
beam as is indicated at the outer fibers in section AA' of Fig. 214. 



At a point in the same section but nearer to the neutral axis, 
as at B, the normal stress on the vertical plane is also found from 
the same expression, in which c is the distance from the neutral 
axis to the point B, but this normal unit-stress is not the maximum 
normal unit-stress at this point since it is combined with shearing 
stresses on the vertical and horizontal planes, the values of which 
are 



(see Art. 40). 


Thus the resulting maximum normal unit-stress at B is 


s' = -|s+jVs 2 +4s s 2 (see equation 202), 


in which s, and hence s', are tensile stresses at the point B. Further, 
the plane on which the stress s' occurs is inclined to the vertical 
cross-section on which s occurs as is indicated in Fig. 214, the 
stress s' is, therefore, frequently called the diagonal tensile unit- 
stress or briefly diagonal tension. 








262 COMBINED NORMAL AND SHEARING STRESSES 


Now in most beams the diagonal tensile unit-stress at any 
point in a cross-section is less than the tensile stress at the outer¬ 
most fiber, and hence need not be found. In reinforced con¬ 
crete beams, however, the diagonal tensile stress is often of impor¬ 
tance since concrete is very weak in tension and, if the beam has 
only longitudinal reinforcing bars in the lower part of the beam, 
the diagonal tension may cause cracking of the beam in that por¬ 
tion of the beam where the shearing stresses are relatively large; 
the diagonal tension crack for the beam loaded as shown in Fig. 
214 would occur along a line similar to XY. 

The relative directions of the maximum normal stresses at 
various points along a section AA' are shown in Fig. 214, and the 
relation of these to the diagonal tension crack XY is also shown. 
In the central portion of the beam where no shear exists the 
cracks are approximately vertical as indicated in Fig. 214. 



Fig. 215.—Directions of resultant stresses in beam. 


\\w //// 


The manner in which the directions of the principal stresses 
vary in a simple beam are shown in Fig. 215; the full lines refer to 
tensile stresses and the dotted lines to compressive stresses. 

In order to prevent failure of concrete 
beams by diagonal tension, some of the 
longitudinal bars are usually bent up at 
the ends as indicated in Fig. 216 so that 
the bars have approximately the same 
direction as the diagonal tensile stresses. 

A diagonal tension failure is fre¬ 
quently called a shear failure and the 
diagonal reinforcing bars are frequently 
called shear bars but it should be kept clearly in mind that the 
failure is due to tensile stresses and the bars are stressed in 
tension. 

PROBLEMS 


Fig. 216.—Steel reinforc¬ 
ing bars for resisting 
diagonal tensile stresses. 


211. The load P acting on the short cantilever beam (Fig. 217) is 800 lb. 
(a) Find the unit-stress at point A normal to the vertical section. (6) Find 








PROBLEMS 


263 


the horizontal and vertical shearing unit-stresses at A. (c) Find the principal 
unit-stresses at A. 



Fig. 217. Fig. 218. 


212. The loads P acting on the simple beam shown in Fig. 218 are each 
equal to 3800 lb. (a) Find the diagonal tensile unit-stress at the point A. 

112. Expression for Ee.—As stated in Art. 22, when principal 
unit-stresses si and S 2 occur on two mutually perpendicular planes 
at a point in a body the maximum unit-deformation, e, at that 
point is in the direction of the larger unit-stress and is equal to 


e 


Si S2 

E m E ’ 


(207) 


in which si is the larger stress and m is Poisson’s ratio (Art. 21). 
When si is a tensile stress e is a unit-elongation, and when si is a 
compressive stress e is a unit-shortening. Further, when S 2 is 
opposite in sign to si it is regarded as negative. 

Now, as shown in Art. 109, when a body is subjected to loads 
that cause shearing stresses on two mutually perpendicular planes 
and a normal stress on one of the planes, there occur principal 
stresses on certain other planes. If, for- example, loads cause a 
tensile stress combined with shear as in the bolt of Fig. 219, the 
maximum unit-deformation is an elongation in the direction of the 
maximum principal stress s' t and its value is 

e= i~ m W’ . (208) 

or 

E e = s' i+ms' c, 

since s' c is negative. 

By substituting in the above equations the values of s' t and 
s' c found in Art. 109 the following equations are obtained: 


Ee = is t +W s t 2 +4s s 2 +m(-|s < +iV s* 2 +4s s 2 ) 
= |(1 -m)s t +i(l+m)Vst 2 +4:S s 2 . 


(209) 



















264 COMBINED NORMAL AND SHEARING STRESSES 


For steel the value of m found from experimental results (see 
Art. 21) is 0.25 to 0.30. Thus, for steel, the above equation 

becomes, _ 

Ee = %s t + fVs 2 *+4s 5 2 , when m=0.25, . . . (210) 

and _ 

£e = 0.35s,+0.65vV+4s s 2 when m = 0.30. . (211) 


Now Ee may be called the equivalent uni-directional unit-stress , 3 
since, if e were caused by a load that developed a stress in one 
direction only (uni-directional stress), the value of Ee would 
then be numerically equal to the actual unit-stress; thus, when a 
bar is pulled by an axial load P in a testing machine, the value of 
the unit-stress, s, is fixed by the conditions of equilibrium and is 
P 

equal to —, but experiment shows that s is also numerically equal 
= E ). Likewise, when two normal stresses at right angles 


to Ee (? 




Fig. 219.—Strain due to principal stresses. 


to each other are developed as in Fig. 219(6) and (c) each stress is 
fixed solely by the conditions of equilibrium but the value of Ee 
is not equal to either of the unit-stresses (see also Art. 22). The 
significance and use of the quantity Ee will be discussed in the 
following article. 

113. Theories of Failure of Elastic Action and Their Applica¬ 
tion.—If a bar is subjected to a gradually increasing axial tensile 
load in one direction only, the material, when the load reaches a 
certain value, will begin to acquire permanent deformation and 

3 Other names, such as simple equivalent stress, strain-stress, etc., have 
been used. However, it is important to note that Ee is hot a real unit-stress 
(force per unit area) in the material but merely a quantity expressed in the 
same units (lb. per sq. in) as a unit-stress. 










THEORIES OF FAILURE OF ELASTIC ACTION 


265 


hence to fail in elastic action, and the reason for the failure may 

be (1) because the normal (tensile) unit-stress — reaches 

the tensile elastic limit, s e (or proportional limit, see Art. 137c), 
or (2) because the shearing unit-stress reaches the shearing elastic 
limit; this requires that the shearing elastic limit shall be equal to 
or less than one-half the tensile elastic limit since, as shown in 
Art. 16, the maximum shearing unit-stress, on a 45-degree oblique 
plane, is one-half the tensile unit-stress, or (3) because the unit- 
deformation reaches a value which cannot be exceeded without 
having part of the deformation plastic and hence permanent, or 
(4) because the energy absorbed by (or the work done on) the 

/I s e 2 1 s 2 \ 

material per unit volume reaches a value (^ “gr °r ^ -gr> see Art. 23 ) 

that can not be exceeded unless the material is given a plastic 
deformation. It is impossible to determine from a simple tension 
test, that is, from a test in which the stress is in one direction 
only, what is the real cause of the beginning of inelastic action in 
a material, because the above limiting values occur simultaneously. 
The four main theories suggested above may be stated briefly as 
follows: 

1. The maximum normal stress theory, often called Rankine’s 
theory, states that elastic action at any point in a material ceases, 
or inelastic action begins, only when the maximum normal (ten¬ 
sile) stress on a certain plane passing through 
the point reaches a value equal to the 
tensile elastic limit found in a simple tension 
test, regardless of the normal or shearing 
stresses that, occur on other planes through 
the point. Thus, if the block in Fig. 220(a) 
reaches its elastic limit when subjected to 
the stress s i, the elastic limit will still be si 
even if the block is subjected to the stress S 2 (Fig. 2205) in 
addition to si. 

2. The maximum strain theory, often called Saint Venant’s 
theory, states that inelastic action at any point in a body due to 
any combination of stresses at the point begins only when the 
maximum unit-elongation e at the point reaches a value equal to 
that which occurs when inelastic action begins in a bar subjected 
to an axial tension test; which is the value of e that occurs simul- 



(a) 


ib) 


Fig. 220. 




266 COMBINED NORMAL AND SHEARING STRESSES 


taneously with the tensile elastic limit s e of the material. Now 
since s e = Ee when the stress is applied in one direction only, the 
maximum strain theory states that inelastic action begins at any 
point in a body when the value of Ee at that point reaches a value 
equal to the tensile elastic limit of the material. 

For example, according to this theory inelastic action in the 
block of Fig. 220(a) begins when si equals the elastic limit of the 
material, but inelastic action in the block of Fig. 220(6) does not be¬ 
gin until Ee in the direction of Si is equal to the elastic limit of the 
material, and this value of Ee will not occur until si in Fig. 220(6) 
is larger than si of Fig. 220(a). For, the unit-deformation in the 
direction si of the block in Fig. 220(a) is made less by the amount 


when the stress S 2 is applied as in Fig. 220(6) (see Art. 22), and 


hence the stress si must be increased before the limiting unit- 
deformation is reached. 

3. The maximum shearing sir ess theory, sometimes called 
Guest’s theory, states that inelastic action at any point in a body 
begins only when the maximum shearing unit-stress on some plane 
through the point reaches a value equal to the shearing elastic 
limit found by testing the material in simple shear as in a torsion 
test (see Fig. 258 and Art. 140). And, as noted above, this theory 
when applied to a simple tension test of a bar requires that the 
shearing elastic limit shall be one-half the tensile elastic limit. 

4. The maximum energy or maximum resilience theory, pro¬ 
posed by Haigh, states that inelastic action at any point in a body 
due to any combination of stresses begins only when the energy 
per unit volume absorbed by the material at that point equals 
the energy per unit volume absorbed by a bar when stressed to 
the elastic limit in a simple tensile test. As shown in Art. 23, the 

| g 2 

value of this maximum energy per unit volume is - 

Z E 


Significance of the Theories. —It must be kept clearly in min 
that the actual unit-stresses at a point in a body are fixed solely 
by the external forces, the stresses being such as to hold the external 
forces in equilibrium, and it is assumed that the stresses are known 
or can be found. The only uncertainty which the theories attempt 
to explain is the state or condition developed in the material that 
causes inelastic action in the material to start. 

Several experimental investigations have been carried out on 


THEORIES OF FAILURE OF ELASTIC ACTION 267 


ductile material, such as steel and brass, with combinations of 
stresses designed to make clear the cause of inelastic action, and 
although the results are not entirely concordant, the best conclu¬ 
sion appears to be that in ductile metals inelastic action begins 
when the shearing unit-stress reaches the shearing elastic limit of 
the material unless the limiting value of Ee is reached before the 
shearing elastic limit is reached, in which case inelastic action 
begins when the limiting value of Ee is developed. The tests 
indicate, however, that the shearing elastic limit is somewhat 
greater than one-half of the tensile elastic limit as demanded by 
the maximum shear theory, the value of the shearing elastic limit 
for steel being close to six-tenths of the tensile elastic limit. The 
maximum energy theory also agrees fairly well with tests results 
for the combinations of stress used in the tests. 

In many of the problems discussed in the preceding chapters 
all these theories lead to the same results. In other problems the 
difference in the results obtained by the four theories are not 
important, and in some problems the assumptions made as to the 
distribution of loads and stresses (see Appendix III) contain 
errors that would not justify strong reliance on the results from 
one theory in preference to those from another. Partly for these 
reasons and partly due to long usage and to the too-frequent 
practice of assuming that the low value of the working stress is 
sufficient to compensate for all uncertainties, the maximum stress 
theory has been most commonly used, although the maximum 
shear theory, for ductile material, has gained rather wide accept¬ 
ance in recent years. The maximum strain theory has been used 
extensively in the design of guns. 

The method of applying the theories to a problem involving 
combined bending and torsion is given below. 

Application of Theories of Failure of Elastic Action .—As stated 
in the preceding article, the dimensions that should be assigned 
to a member which is subjected to loads causing combined stresses 
depend on the theory held concerning the cause of the breakdown 
of elastic action. This fact will be illustrated in the solution of a 
problem of the design of a member subjected to loads that cause 
the combination of stresses discussed in Art. 109; namely, shearing 
stresses on two planes at right angles to each other and a normal 
stress on one of the planes. In the following problem this com¬ 
bination of stresses is caused by combined bending and torsion. 


268 COMBINED NORMAL AND SHEARING STRESSES 


ILLUSTRATIVE PROBLEM 

Problem 213. The crank-shaft shown in Fig. 208 is made of steel having 
a tensile and compiessive proportional limit of 40,000 lb. per sq. in. Deter¬ 
mine, by each of the four theories of failure, the diameter, d, the shaft should 
have to resist the load P without developing more than one-half of the max¬ 
imum elastic strength of the material. Tests show that the shearing propor¬ 
tional limit of steel is approximately 0.6 of the tensile proportional limit. 

Solution. —As noted in Problem 202, the twisting moment T and the bend¬ 
ing moment M have the following values: 


T = 6P = 60,000 lb .-in. and M = 8P = 80,000 lb .-in.; 
and the shearing and normal unit-stresses due, respectively, to T and M are 
Tc 167 7 , Me 32 M 

and S= T"• 


(а) Maximum Normal Stress Theory. —The maximum normal unit-stress 
resulting from the unit-stresses s s and s, as shown in Art. 109, is 

s' = is+|Vs 2 +4s a 2 , 

and according to the maximum stress theory inelastic action in the material 
begins when s' exceeds 40,000 lb. per sq. in Hence the working unit-stress 
is 20,000 lb per sq. in. Therefore, 

20 ,ooo J- m+iJirnTJ *™ry 

2 7rd 3 2 \ \ ird 3 / \ ird 3 ) 

=^| (M+VlHT), 

-ird 3 

whence 

= ^oo (80,00 ° +V (80,000) ’ +(60 - 000 > 2 >> 

and 

g _ 

d 3 =-( 8 +V(8) 2 +(6) 2 ) =45.8 in. 3 , 

7r 

Therefore 

d = 3.57 in. 

(б) Maximum Strain Theory —According to the maximum strain theory, 
inelastic action in the material begins when the maximum value of Ee becomes 
40,000 lb. per sq. in. And, from Art. 112 the maximum value of Ee, assuming 
m equal to 0.30, is given by the equation 


Pe = 0.35s+0.65V / s 2 +4s 8 2 . 










ILLUSTRATIVE PROBLEM 


269 


The working value of Ee is \ X40,000. Thus, 


Hence 


Therefore 


20,000 = — 7 - (0.35M+0.65Vilf 2 +T 2 ). 

ird 3 

d 3 = —(0.35 X8+0.65X10) = 47.4 in. 2 

TV 

d = 3.62 in. 


(c) Maximum Shearing Stress Theory .—According to this theory inelastic 
action in the shaft begins when the maximum shearing unit-stress s' Sf as given 
by the equation 


s' s = |V S 2+4s fi 2 , 


reaches the shearing proportional limit, that is, 0.6X40,000 or 24,000 lb. per 
sq. in. The working unit-stress then is 12,000 lb. per sq. in. Hence, 

12,000 = iVs 2 +4s s 2 


Hence, 


Therefore, 


= -— Vm 2 +T 2 . 

ird 3 


160 /- 1600 

d 3 =—V8 2 +6 2 =-; 

V2ir \2ir 


42.5 in. 3 


d = 3.48 in. 


(d) Maximum Energy Theory .—As shown in Art. 23 the expression for the 
energy absorbed by the material per cubic inch when subjected to the stresses 
s and s s , as in the shaft here considered, is 


Is 2 1 s s 2 
2 E + 2 ¥ s ’ 


But, for steel E 8 = %E (Art. 5) and hence the above expression may be written 


1 sf H 2 

2 E + 4 E' 


Now, according to the maximum energy theory the maximum value of the 
energy U that can be absorbed per unit volume without causing inelastic 
action in the material is 


2 E 


1 (40,000) 2 

2 30,000,000 


= 26.6 in.-lb. per cu. in. 








270 COMBINED NORMAL AND SHEARING STRESSES 


But since it is specified that the material shall be required to absorb \ X26.6 — 
13.3 in .-lb. per cu. in., when the shaft is subjected to the given load, we have, 


Is 2 5 s, 2 

13 - 3 = -2^ + 4lP 
4Xl3.3X£ = 2s 2 +5s* 2 


/32M \ 2 , /16 T 

=2 \~^) + 5 t< 


M 3 ) 2 =11,100, 


d = 3.22 in. 


Thus, in this problem the maximum strain theory requires the largest shaft. 


PROBLEMS 

214. In the above problem change the length of the crank from 6 in. to 
12 in. and find, by each of the four theories, the diameter of the shaft required. 

215. Find, by all four theories of failure of elastic action, the area of 
cross-section that the steel bolt shown in Fig. 219(a) should have in order to 
resist a load P of 4000 lb. and a load Q of 3000 lb. without developing more 
than \ of the maximum elastic strength of the material. The bolt is made of 
steel having a tensile proportional limit of 30,000 lb. per sq. in. Use a value 
of 0.25 for Poisson’s ratio. 

216. Find, by all four theories of failure of elastic action, the diameter of 
the shaft shown in Fig. 202, if the load P is 8000 lb. and the torsional moment, 
Qq, is 16,000 lb.-in. Assume that the shaft, in resisting the loads, develops 
0.65 of the maximum elastic strength of the material. The shaft is made of 
steel having a tensile elastic limit of 60,000 lb. per sq. in. Use a value of 0.25 
for Poisson’s ratio. 

217. Prove that, when a cylindrical shaft is subjected to combined bending 
and torsion, the ratio of the maximum shearing stress s' a to the shearing 
yield point will be equal to the ratio of Ee to the tensile yield-point when 
the ratio of the twisting moment, T, to the bending moment, M, is 
approximately equal to 1.62, provided that the shearing elastic limit is 0.6 
of the tensile elastic limit. (See Problem 207 for the ratio of T to M accord¬ 
ing to the maximum stress theory.) 

218. A circular steel shaft having a tensile elastic limit of 36,000 lb. per 
sq. in. is subjected to a bending moment of 50 ton-in. and a twisting moment 



PROBLEMS 


271 


of 50 ton-in. Assume that the shearing elastic limit is 0.6 of the tensile 
elastic limit, (a) Compute by each of the four theories of failure of elastic 
action the diameter the shaft must have if it develops § the maximum elastic 
strength of the material. ( b ) Change the bending moment to 5 ton-in. and 
repeat the problem, (c) Change the twisting moment to 5 ton-in. leaving the 
bending moment at 50 ton-in. and again repeat. 

Ans. (a) 4.68; 4.74; 4.65; 3.97. ( b ) 3.60; 3.84; 4.14; 3.39. (c) 4.40; 

4.40; 4.14; 3.67. 


CHAPTER XIII 


IMPACT AND ENERGY LOADS 

114. Introduction.—In the preceding chapters, the loads 
acting on the members were assumed to be gradually applied 
(static loads). To state the same idea in other words, the bodies 
that applied the loads were not in motion when they came in con¬ 
tact with the resisting member, and hence they delivered no 
kinetic energy to the resisting member. 

Members of engineering structures and machines, however, 
frequently must resist loads that are applied by moving bodies, 
and the kinetic energy of these moving bodies must be absorbed 
by the resisting member, thereby developing stresses and defor¬ 
mations in the member. 

Now there are two methods of determining the stresses and 
deformations developed in the resisting member: (1) To estimate 
the maximum pressure or force exerted by the moving body on 
the resisting member; this force is then considered to be a static 
load and is used in the equations developed in the preceding chap¬ 
ters. (2) To estimate the energy that is absorbed by the resisting 
member and from this value of the energy determine the stresses 
and deformations by use of the equations developed in this chapter. 

In the first method the load is a force and is, of course, expressed 
in pounds, tons, etc.; it is called an impact load and the method 
is frequently called the equivalent-static-load method. Whereas, 
in the second method the load is considered to be a quantity of 
energy and is therefore called an energy load; it is expressed in 
foot-pounds, inch-tons, etc. 

The determination of the stresses in structural and machine 
members that are subjected to impact or energy loads is less 
definite and satisfactory than in members subjected to static loads, 
for the reason, that, in general, the uncertainty as to the value of 
an impact load or of an energy load is greater than that of a static 
load. The design of members subjected to energy loads, there- 

272 


CALCULATION OF ENERGY 


273 


fore, is based more directly on the study of existing machines and 
structures than is the design of members subjected to static loads. 
However, in this connection it is well to recognize, as stated in 
Art. 7, that the static loads to which members in structures and 
machines are assumed to be subjected frequently are only rough 
approximations of the actual loads that the members are required 
to resist. 

When the energy delivered to the resisting member is relatively 
small the equivalent-static-load method is the one more commonly 
used in determining the stresses in the member. Thus the load 
caused by a moving train on a bridge, by wind on a building, by a 
moving crowd of people on a stadium, etc., is assumed to be 
equivalent to a static load composed of the actual static or “ dead ” 
load plus a static load assumed to be equivalent (in producing 
stress) to the dynamic or impact effect of the load. This addi¬ 
tional (equivalent) static load is frequently called the “ live ” 
load. 

If, on the other hand, the energy delivered to the resisting 
member is relatively large the energy-load method may be the 
more useful one, for, it will be found that the dimensions of the 
resisting member and the properties of the material in the member 
that give it maximum resistance to an energy load are quite dif¬ 
ferent from those that give the member maximum resistance to a 
static load. 

115. Calculation of Energy Delivered to Resisting Member.— 

As a rule, only a part of the energy of the moving body that 
delivers the energy to the resisting member is absorbed or stored 
in the resisting member; some of the energy is spent in each of the 
following ways: 

(а) In causing stresses and deformations throughout the 
moving body itself. 

(б) In causing local stresses and deformation of both bodies 
at the surface of contact, especially if the velocity of the moving 
body is large when it comes in contact with the resisting body. 

(c) In overcoming the inertia resistance of the resisting mem¬ 
ber, and 

(i d ) In deforming and in moving the external supports to which 
the resisting member is attached. 

Thus, for example, an airplane when landing possesses kinetic 
energy which is delivered mainly to the wheels and axle on which 


274 


IMPACT AND ENERGY LOADS 


the wheels are mounted, and these parts are constructed so that 
they can absorb a large amount of the energy. But, a consider¬ 
able part of the energy of the airplane is absorbed as stated under 
(a) and (d) above, that is, in causing deformation of the frame of 
the airplane and in deforming the tires and spokes of the wheels 
and in causing depressions in the ground. 

Kinetic Energy of Bodies .—If the moving body that delivers 
the energy has a motion of translation or of rotation the kinetic 
energy of the body may be found from the expressions given below; 
from this kinetic energy must be subtracted the amount of energy 
assumed to be lost, in order to obtain the energy load absorbed 
by the resisting member. 

Translation .—The kinetic energy E k , of a body having a motion 
of translation is 

E h = -Mv 2 = l^v 2 .( 212 ) 

2 2 g 


in which v is the velocity of the body and M is the number of units 
of mass in the body (found by dividing the weight, W, of the 
body by g, the acceleration which the weight causes when it is 

the only force acting on the body = — J. When W is expressed 

in pounds, g in feet per second per second, and v in feet per second, 
E k will be expressed in foot-pounds. Further, if the body acquires 
the velocity v by falling freely from rest through a distance of h 
feet, then v 2 is equal to 2 gh and hence E k =Wh. (The value of g 
is 32.2 ft. per sec. 2 , approximately). 

Rotating .—If the body that delivers the energy load has a 
motion of rotation about a fixed axis its kinetic energy E k may be 
found from the expression 

E k = il 0 u 2 , .(213) 


in which co is the angular velocity of the body and T 0 is the moment 
of inertia of the body about the axis of rotation; I 0 is expressed 
/W\ 

in terms of the mass of the body and the dimensions of the 

body. Thus, for a solid cylinder that rotates about its geometric 
1 W 

axis /o = o— r 2 , where r is the radius of the cylinder. If IF is 
^ 0 

expressed in pounds, g in feet per second per second, the dimen- 


STRESS IN BAR DUE TO AXIAL ENERGY LOAD 275 


sions of the body in feet, and oo in radians per second, E k will be 
expressed in foot-pounds. The values of Jo for bodies of other 
shapes may be found in books on analytical mechanics and in 
engineers’ handbooks. 

116. Stress in Bar Due to Axial Energy Load.—Let an energy 
load U be applied to a bar so that the bar is subjected to an axial 
tensile stress. Such a load occurs in machines of various types 
as will be discussed later; the method of estimating the value of U 
is given in the preceding article; namely, the value of U is the 
kinetic energy of the moving body less the part of this energy 
that is lost in ways stated in Art. 115. 

The problem is to find the maximum unit-stress in the bar 
caused by the energy load U. The assumption 1 is made that a 
material when resisting an energy load acts in the same way as 
when resisting a gradually applied (static) load; namely, unit- 
stress is proportional to unit-deformation until the proportional 
limit is reached. Hence, the energy U stored in the bar when the 
unit-stress in the bar is s (which is equal to the work, w, done on 
the bar in causing the stress s, provided that s is not greater than 
the proportional limit) may be expressed as follows: 

XJ=w = %Pe, 


in which P is the final value of the gradually applied load and e is 
the total elongation of the bar. But since s is within the propor¬ 


tional limit we have -=E. 

e 


Further, s = — where a is the cross¬ 
es 


sectional area of the bar. Hence, 


TT 1 S 2 7 

U= 2 E al 


Now, since the energy stored in the bar is assumed to be inde¬ 
pendent of the way the energy is delivered to the bar, the rela¬ 
tion between any energy load U and the unit-stress s is 


TT 1 S 2 7 

U = 2E al 


4 


2 UE 
al ’ 


(214) 


1 This assumption probably is not true if the load is applied with extreme 
suddenness and acts on the body only a very short period of time, causing a 
“ momentary ” stress. But with energy loads as they usually occur in engi¬ 
neering practice the assumption is probably justified although experimental 
verification is difficult to obtain. 


276 


IMPACT AND ENERGY LOADS 


in which s must not exceed the proportional limit of the material. 
If U is expressed in inch-pounds, E in pounds per square inch, a in 
square inches, and l in inches, s will be expressed in pounds per 
square inch. Equation (214) shows that the energy absorbed 
per unit-volume of the material, when stressed to the proportional 
Is 2 

limit, s ey is •= this value is represented by the area OCD in Fig. 
z Jti 

221 and is called the modulus of resilience of the material (see 
Art. 23 and 146). 



The ideal material, then, for resisting energy loads in service 
in which the material must not incur permanent distortion, is one 
having a high modulus of resilience, that is, one having a high 
proportional limit and a low modulus of elasticity (see Art. 146 for 
further discussion). 

Ultimate Energy Resistance of a Material. Toughness. —The 
ultimate energy resistance of a material is the maximum amount 
of work that can be done on the material per unit of volume* 
without causing the material to rupture. It is represented by the 
total area under the stress-strain curve, 1 area OCEF in Fig. 221. 
Most of this work is dissipated in heat and in causing permanent 
deformation of the material, and only a very small part is stored 

1 It is probable that the ultimate energy resistance is greater than that 
represented by the area under the static stress-strain curve, and hence the 
error introduced is on the side of safety. 






















COMPARISON OF STATIC AND ENERGY LOADS 277 


in the material as stress-energy that can be recovered when the 
stress is released. The property of a body by virtue of which 
work can be done on the body when stressed beyond its elastic 
limit is called toughness. A tough material is needed, therefore, to 
resist energy loads when the material in service is likely to be 
stressed beyond its yield-point. 

For materials having a stress-strain diagram similar to that 
shown in Fig. 221 (that is, for ductile materials) the area under the 

curve is given approximately by the expression and hence 

the ultimate energy load U the bar can resist is, approximately, 


U 



Fig. 222.—Stress-strain diagram 
brittle material. 


*±VaJ,.(215)' 

in which s y and s u are the yield- 
point and ultimate strength of 
the material, respectively, and e u 
is the ultimate unit-deforma¬ 
tion. And, for material having 
a parabolic stress-strain diagram 
(Fig. 222) such as cast iron and 
concrete, the area under the 
curve, is approximately f s u e u , 
and hence the ultimate energy 
load a bar of such a material can 
resist is, approximately, 

U = is u e u .(216) 


117. Comparison of Effects of Static and Energy Loads.— 

The resistance offered by a bar to a static axial load, P, depends 
only on the maximum unit-stress developed, which occurs on the 
smallest cross-section (P = as ), whereas the resistance of the bar 
to an energy load, U, as indicated by equation (214), depends not 
only on the maximum unit-stress, s, but also (1) on the distribu¬ 
tion of the stress throughout the body, since the energy absorbed 

1 

by a given unit of volume is - and hence depends on the degree 

z hi 


to which that volume is stressed, and (2) on the number of units 
of volume ( al ) of material in the bar. The influence of these two 
factors may be shown as follows: 










278 


IMPACT AND ENERGY LOADS 


(1) The static strengths of the two cylindrical bars shown in 
Fig. 223 when subjected to axial loads are equal, since the smallest 
cross-sections are equal, and hence the loads P required to pro¬ 
duce a given stress s in the bars are equal (P = as ). The energy 
loads required to produce a given stress 
in the two bars, however, are very dif¬ 
ferent; the bar having the constant 
diameter, and hence a uniform distribu¬ 
tion of stress throughout the length of 
the bar, is able to absorb the greater 
amount of energy. For example, if 
the diameter of the upper hah of the 
bar shown in Fig. 223(6) is reduced 
from 2d to d the area a (and also 
volume al) will, thereby, be decreased 
to one-fourth of the original area (and 
volume), and hence the unit-stress will 
be increased to four times the original 
value. Therefore, the energy absorbed 



per unit volume (- 


(IS) by 


this upper 


half will be sixteen times as great as 

„ ™ ™ i , that absorbed per unit-volume when the 

Fig. 223.—The bars have • oj to. • 

equal resistance to static diameter is 2d. Thus, in the expres- 

1 s 1 2 

sion - for the upper half of the 

2 hi 


loads, but not to energy 
loads. 


1 s 2 

bar, the factor - -= has been increased more than the factor al 

2 h 


has been decreased; the total energy absorbed by the upper half 
of the bar being increased to four times its original value by 
reducing the diameter of the upper half from 2d to d. 

This method of increasing the energy load that can be applied 
to a bar was brought forcibly to the attention of engineers in the 
early development of the Straight Line engine by Professor John 
E. Sweet. The bolts in the connecting-rod head when made in 
the usual form with full-sized shanks and threaded ends as in 
Fig. 224(a) frequently broke 3 in service due to the energy load 


3 As will be shown in the next chapter, failure of material is frequently 
due to the repeated application of a stress if the stress is above a limiting 
minimum value, and this was probably the cause of the failure of the bolts 























COMPARISON OF STATIC AND ENERGY LOADS 279 


delivered by the reciprocating parts of the engine. By turning down 
the bolts so that the area of cross-section of the shank was equal to 
that at the roots of the threads, as shown in Fig. 224(6), the dif¬ 
ficulty was removed since most of the energy of the reciprocating 
parts was then absorbed in the shank of the bolt, and the amount 



(a) 



Fig. 224.—Energy resistance of bolt increased by decreasing diameter of 

shank. 


absorbed at the roots of the thread was thereby greatly reduced 
with a corresponding reduction in the stress developed at the roots 
of the threads. 

It will be seen, therefore, that uneven distribution of stress 
throughout a member may greatly reduce the energy load that 
can be applied to the body. 

2. If the stress in a bar is distributed uniformly throughout 
the bar, that is, if the area of cross-section is constant, the energy 
load that the bar will resist may be increased by increasing the 
length of the bar since this increases the volume, each unit of vol¬ 


ume absorbing the same amount 



of energy. 


This method of increasing the energy resistance of a member 
was clearly illustrated in the early development of rock drills. 4 
The cylinder heads of the rock drills were attached to the cylinder 
as shown in Fig. 225(a) and since in the operation of the drill it is 
impossible to avoid occasional sharp blows of the piston on the 
cylinder head the bolts were subjected to severe energy loads and, 
as a consequence, broke. The trouble was remedied by using 


here referred to. But, by reducing the stress in the bolts, as stated, the rep¬ 
etitions of this lower stress did not cause failure. 

4 Halsey, F. A. “Materials and Constructions for Resisting Shock,” 
American Machinists, Sept. 9, 1915, p. 459. 































280 


IMPACT AND ENERGY LOADS 


bolts long enough to extend through both cover plates as shown 
in Fig. 225(6). 



(a) (6) 

Fig. 225.—Energy resistance of bolts increased by making them longer. 


118. Special Cases of Axial Energy Loads. —(a) Sudden Load .— 
A sudden load is a force that remains constant throughout the 
entire deformation, e, of the resisting member. Thus, if a sudden 
axial load P act on a bar the corresponding energy load, U t sup¬ 
plied to the bar (that is, the work done on the bar in causing the 
deformation e) is Pe. 

Thus the expression for s in equation (214) becomes, 


-4 


2 PeE 
al ’ 


(217) 


and since j = e, and Ee = s, this expression may be written 

P P 

s 2 = 2—s or s = 2 —. . . . 

a a 


(218) 


But if the load P were a static (gradually applied) load, — would 

be the unit-stress developed. Therefore, within the propor¬ 
tional limit, the unit-stress caused by a sudden load is twice as 
great as that caused by the same load when gradually applied. 

(6) Energy Load Due to a Falling Body .—If the energy load is 
delivered to a bar by a body of weight W falling through a height 
h before it comes in contact with the bar (see Fig. 229), the value 
of U in equation (214) is W(h-\-e), where e is the total elongation 
of the bar. Hence, 


4 


2W( h+e)E 
al 


(219) 


But since e depends on s it will be convenient to express e in terms 







































SPECIAL CASES OF AXIAL ENERGY LOADS 


281 


of s, and at the same time to express s in terms of the unit-stress, 
si, and deformation, ei, that would be caused by the same weight 
if gradually applied. Thus, the above equation may be written: 


2 _2 WE(h+e) 
al 


But — equals the unit-stress, si, due to the load W when grad¬ 
ually applied. Further, within the proportional limit, s = Ee; 
also e — el, and hence y=-• Therefore the above equation may be 

i 6 


written: 

S =2 Sl g+ 1 ), 


Now the ratio of unit-stress to unit-deformation is assumed to be 
constant within the proportional limit regardless of the kind of 
load that causes the stress and deformation. Hence, 


S_S\ 

€ ei 

Also, - =—, since e = el and ei = eih, 
e ei ■ 

in which si is the unit-stress and e\ the total deformation 
that would be caused by a static load W. Thus e =—- and hence 

.= 

\se 1 / 

Whence, the maximum stress is, 

s = si+sia/h-, . . . . . (220) 

\ ei 


in which s must not exceed the proportional limit. This equation 
shows that the stress due to an energy load caused by a falling 
weight may be greatly in excess of that caused by the same weight 
when gradually applied. 

Similarly, since 4 = e and € 1 , the elongation e of the bar due 

Jh hi 


to the falling weight is 


e = ci+eix/l + 


4- 


2h 
ei 


( 221 ) 







282 


IMPACT AND ENERGY LOADS 


in which e\ is the elongation of the bar that would be caused by the 


same weight if gradually applied. 

119. Working Stress and Working Value of Energy. —If the 

equivalent-static-load method (see Art. 114) is used in making 
allowance for the effect of the impact of the load, the working 
stress may be the same with impact loading as with static loading. 
Specifications usually require, however, slightly better material 
(more uniform in quality and tougher) for use in structures sub¬ 
jected to impact, such as, ships, locomotives frames and bridges, 
than for structures subjected to loads that are approximately static 
loads, such as buildings. 

If, on the other hand, the load is considered to be- an energy 
load, the working stress, or rather the working value of the energy, 



should be based on the modulus of resilience 


rial, particularly if the maximum usable energy of the material is 
the elastic energy, that is, if the resisting member would be struc¬ 
turally damaged if stressed above the proportional limit (or yield 
point) of the material. But the toughness of the material should 
also be considered in order to insure reserve energy for resisting 
overloads and high localized stresses (see Art. 147 and 148 for 
further discussion). 

A working value of energy should, of course, be considerably 
less than the modulus of resilience; the working value depends on 
the way the energy load is applied, on the service rendered by the 
member or machine, on the toughness of the material (its capacity 
to resist overloads, etc.), and on the form of the member with 
reference particularly to the way the energy is distributed through¬ 
out the member. In well designed machines (in which the energy 
load on any member is reduced to a minimum by the balancing 
of moving parts, by the use of springs, by adjusting the rela¬ 
tive stiffness of the component parts and of the connected mem¬ 
bers, etc.), the working value of the energy may be made one- 
fifth to one-tenth of the modulus of resilience. It should be noted 
that the resilience varies as the square of the stress developed 
(assuming E to be constant as in the case of steel), and hence 
when the energy absorbed by the member is reduced a given 
amount the corresponding stress is reduced considerably less than 
this amount. 

Table 5 should be helpful in selecting material for resisting 


ILLUSTRATIVE PROBLEMS 


283 


energy loads and in selecting working values of energy to use in 
proportioning members subjected to energy loads. 

TABLE 5 

Average Values of Modulus of Resilience and Toughness 


Material 

Tensile 
Propor¬ 
tional 
Limit 
(lb. per 
sq. in.) 
se 

Tensile 
Ultimate 
Strength 
(lb. per 
sq. in.) 

su 

Tensile 
Modulus 
of Elastic¬ 
ity 

(lb. per 
sq. in.) 

E 

Ultimate 
Elonga¬ 
tion Per 
Inch of 
Length 
(in.) 

Cm 

Tensile 
Modulus 
of Resili¬ 
ence (in.-lb. 
per cu. in.) 

1 s e 2 

2 E 

Toughness 
in Tension 
(Represented 
by area under 
Stress-strain 
Diagram) 
(in.-lb. per 
cu. in.) 

Low carbon steel. 

30,000 

60,000 

30,000,000 

0.35 

15.0 

15,700 

Medium carbon steel.. 

45,000 

85,000 

30,000,000 

0.25 

33.7 

16,300 

High carbon steel. 

Special alloy steel 

75,000 

120,000 

30,000,000 

0.08 

94.0 

5,100 

(Heat treated). 

200,000 

230,000 

30,000,000 

0.12 

667.0 

22,000 

Gray cast iron. 

6,000 

20,000 

15,000,000 

0.005 

1.2 

70 

Malleable cast iron. . . 

20,000 

50,000 

23,000,000 

0.10 

17.4 

3,800 

Rolled bronze. 

Timber (Hickory) .... 

40,000 

5,500* 

65,000 

10,000* 

14,000,000 

2,400,000* 

0.30 

57.2 

6.32* 

15,500 


* In compression. The effectiveness of timber for resisting energy loads is greater than 
that indicated by the value of the resilience given in the table, due in part to the fact that 
timber may be stressed somewhat above its proportional limit without destroying its use¬ 
fulness; that is, the properties of resilience and toughness are not sharply defined. 


ILLUSTRATIVE PROBLEMS 

Problem 219. Fig. 226 shows two bolts with square threads; they have 
the same dimensions except that one has the shank turned down to a diameter 
equal to that at the root of the threads for a length of 10 in. If both bolts 



Fig. 226. 


are made of soft steel with a tensile proportionality limit of 32,000 lb. per 
sq. in., find the axial static load and the axial energy load that each bolt 
will resist when stressed to the proportionality limit. 















































284 


IMPACT AND ENERGY LOADS 


Solution.—Static Loads. —The static loads P that the two bolts can resist 
are equal since the least cro&s-sectional areas are equal. Thus, 

P = as =^-^^-32,000 = 2,265 lb. 

4 

Energy Loads. —The cross-sectional area, a\, of each bolt at the root of the 
threads is 0.0706 sq. in. and a 2 of the shank is 0.1963 sq. in. Hence when the 
unit-stress $i in the first inch of the bolt (the threads are neglected) is 32,000 
lb. per sq. in. the unit-stress s 2 in the shank of the bolt is only 

^-^-^X32,000 = 11,520 lb. per. sq. in. 

0.1963 ’ ’ 

Therefore, the energy load U that can be absorbed by the bolt in Fig. 226(a) 
when stressed to 32,000 lb. per sq. in. is 


1 si 2 1 s 2 2 

U = 2j aih+ 2¥ a2h 


1 (32,000) 2 

2 30,000,000 


0.0706X1 + 


1 (11,520)2 

2 30,000,000 


X0.1963X17 


= 1.21+7.38 = 8.59 in.-lb. or 0.715 ft.-lb. 


Likewise, the maximum energy load that can be applied to the bolt in Fig. 
226(6) without causing a stress greater than the proportional limit is 


TJ 1 Sl2 7 i 1 *2 2 7 

U = 2~E a ' ll+ 2j a ‘ h 


1 (32,000)2 

2 30,000,000 


0.0706X11 + 


1 (11,520)2 

2 20,000,000 


0.1963X7 


= 13.3+3.05 = 16.4 in.-lb. or 1.36 ft.-lb. 


Therefore, a body weighing only 0.715 lb. falling a distance of only 1 ft. 
would supply enough energy to the bolt in Fig. 226(a) to stress it to the pro¬ 
portional limit provided all the energy were absorbed by the bolt, whereas a body 
weighing 2265 lb. would be required to cause the same stress if its weight were 
gradually applied to the bolt. Further, turning down the shank of the bolt 
as shown in Fig. 226(6) nearly doubles the energy resistance of the bolt without 
affecting its static resistance. 

The results found in this problem further emphasize the importance of a 
uniform distribution of stress throughout a body that is required to resist 
energy loads. The results also indicate that energy loads would be extremely 
serious if the energy did not distribute itself throughout the whole structure 
or machine, or if it did not dissipate itself in the ways stated in Art. 115. 
These facts are further emphasized in the next problem. 








ILLUSTRATIVE PROBLEMS 


285 


Problem 220. In Fig. 227 is shown a punching machine. The fly wheel A is 
made to rotate by means of friction discs (not shown); it supplies the energy 
required to punch a f-in. hole through a f-in. steel plate. The diameter of the 
screw at the roots of the threads is 4 in. and the length of the screw beneath 
the frame is 10 in. 

(a) Find the maximum compressive unit-stress in the stem of the screw 
assuming that the load is a static axial load. (6) Find the maximum unit- 
stress, assuming that the load is an energy load and that the screw absorbs 
all the energy, (c) Find same as under (6), assuming that the machine frame 
absorbs three-fourths of the energy and the screw the other fourth. 




Solution. —(a) The maximum force P exerted on the punch, if the ulti¬ 
mate shearing unit-stress of the plate is 60,000 lb. per sq. in., is 

P = shearing area X 60,000 

= Trd X t X 60,000 = 7r| X f X60,000 

= 103,000 lb. 

And the unit-stress in the screw due to a static load of 103,000 lb. would be 


P 




a 


103,000 

7t(2)* 


= 8210 lb. per sq. in. 


(6) Tests show that the work diagram for the punching of steel plates is 
approximately that shown by the heavy line in Fig 228, and that the area 
under the work diagram is equal, approximately, to the area under a triangular 
diagram (Fig. 228) in which the maximum pressure is that required to develop 
a shearing unit-stress of 60,000 lb. per sq. in. in the steel in order to punch 











































286 


IMPACT AND ENERGY LOADS 


the hole. Hence, the work done in punching the hole as represented by the 
area under the triangular diagram is 

P 103,000 5 

y= 2 X< = -t- X 8 

= 32,200 in.-lb. 

If all of this work were stored in the part of the screw between the plate 
and the machine frame and none in the machine frame, the unit-stress in 
the screw would be 

WE 1 2X32,200X30,000,000 
~aT~\l 12.56X10 

= 124,000 lb. per sq. in. 

provided that this value is less than the proportional limit. 

(c) But the frame is not rigid, and hence probably a considerable part of 
the work done by the plate on the punch and screw is stored in the frame of 
the machine. If three-fourths of the work is used in deforming the frame 
then the value for U in the above expression is 8050 in.-lb. and the unit-stress 
becomes 

s = 62,000 lb. per sq. in. 



It is clear from the above results that a large part of the energy delivered 
to the screw is transmitted on to the machine frame. And, in general, the 
distribution of energy throughout the whole machine insures the safety of the 
machine; consideration of this fact is of great importance in the design of 
various types of machines and structures. 

The results of this problem also suggest the desir¬ 
ability of a study of existing machines when designing 
members that are subjected to energy loads. 



PROBLEMS 

221. A body A, Fig. 229, having a weight of 20 lb. 
falls a height A of 4 ft. when it comes in contact with 
the end of a rod having a length l of 6 ft. (a) If the 
rod is made of soft steel having a proportional limit 
of 30,000 lb. per sq. in. what cross-sectional area 
should the rod have in order to prevent the unit- 
stress from exceeding the proportional limit? ( b ) If 
test specimens of the steel when tested had an ulti¬ 
mate strength of 60,000 lb. per sq. in. and an elongation 
of 30 per cent estimate the weight of A required to 
rupture a rod having a diameter of 0.25 in. (Note: 
All the energy of the falling weight is assumed to be 
absorbed by the rod.) 

Ans. (a) a = 0.897 sq. in.; (6) IF = 684 lb. 























PROBLEMS 


287 


222 . If the rod in Problem 223 were a stick of oak having a proportional 

limit of 3500 lb. per sq. in. and a modulus of elasticity of 1,800,000 lb. per sq.in., 
what should be the least cross-section of the stick to prevent the stress from 
exceeding the proportional limit? Ans. a = 3.92 sq. in. 

223 . The piston rods of the forging hammers in a certain steel plant fre¬ 
quently broke, and when they were finally replaced by a higher carbon steel 
the trouble ceased. If the proportional limit of the soft steel was 32,000 lb. 
per sq. in. and that of the higher carbon steel was 50,000 lb. per sq. in., how 
much more energy could the higher carbon steel rod absorb without being 
stressed above the proportional limit? Note: The breaking of the rods was 
probably due to the fact that the stress was repeated (repeated stress is dis¬ 
cussed in the next chapter), but this was the secondary cause of the failure. 
(See footnote 2 of Art. 117.) 



0 0.02 0.04 0 0.002 0.004 

Unit-deformation 


Fig. 230.—Stress-strain diagrams of two grades of steel. 

224 . Fig. 230 shows stress-strain diagrams for medium carbon steel, 
(about 0.30 per cent carbon) and for high carbon steel (about 0.80 per cent 
carbon), (a) If a bolt of medium carbon steel is stressed just to its propor¬ 
tional limit by an energy load of 25 foot-pounds, how large an energy load 
will a bolt of the same size of the 0.80 per cent carbon steel resist when stressed 
to its proportional limit? (6) What will be the ratio of the energy loads that 
will cause rupture of the two bolts? Ans. (a) 56.2 ft. lb.; ( b ) 6.5 approx. 

225 . In Fig. 231 are shown stress-strain diagrams for cast iron and for 

steel castings having about 0.20 per cent of carbon. Owing to the greater 
strength of steel the cross-sectional area of a steel casting for railway service 
(car couplers for example) is made only 0.4 as great as that of a cast-iron cast¬ 
ing. What will be the relative resistance to rupture, when subjected to an 
axial energy load, of the steel coupler and the cast-iron coupler if the lengths 
are the same? Since the service of a car coupler is not destroyed by perma¬ 
nent deformations and since in railway service such deformations are very 
likely to occur, the ultimate rather than the elastic energy loads are the gov¬ 
erning factors. Ans. 18.4 times as great. 

226 . A tension member in a certain machine is subjected to an energy 
load. If the member can be given a permanent deformation without destroy¬ 
ing its usefulness and can be made of either cast iron or oak, which material 




























288 


IMPACT AND ENERGY LOADS 


should be used provided that the strength of the member is the only factor to 
be considered (make use of Fig. 231)? Ans. Oak. 



0.01 0.02 0.03 0.04 0.05 0.06 

Uni t-deformation 


Fig. 231.—Stress-strain diagrams. 

227. A bar 3 ft. in length and \ sq. in. in cross-sectional area is stressed 
to its proportional limit by an axial energy load U. If the diameter of the 
bar is turned down to one-half of its original value over one-fourth of its 
length, what is the value of the energy load that will stress the bar to its pro¬ 
portional limit? 

120. Stresses in Beams Due to Energy Loads. —The amount of 
energy that a beam will absorb when stressed within the propor¬ 
tional limit of the material depends on the type of beam (canti¬ 
lever, simple, fixed, etc.) and on the type of load (concentrated, 
distributed, etc.) in addition to the form and dimensions of the 
beam. 

Simple Beam with Concentrated Load at Center .—Let it be 
required to find the maximum stress caused by an energy load U 
applied at the center of a simple beam, when the stress in the beam 
does not exceed the proportional limit of the material. Now the 
energy that can be stored in a beam is assumed to be the same when 
subjected to a gradually applied load as when subjected to an 
energy load due to a moving body. Thus the expression for the 
energy load U that can be absorbed by the beam when the stress 
does not exceed the proportional limit may be found as follows: 

Let a simple beam having a span length of l feet be loaded with 
a gradually applied concentrated load at the mid-span, the final 
value of which is P pounds. The relation between the load P, 


























STRESSES IN BEAMS DUE TO ENERGY LOADS 289 


and the deflection, A, of the beam is shown by the curve in Fig. 
232(6). The energy load U that can be absorbed by the beam 
(which is equal to the work done on the beam as a gradually 
applied load increases from zero to the value P (represented hy 




Fig. 232.—Energy resistance of beam; load at center. 


AB in Fig. 2326) is represented by the area OB A, and may be 
found from the expression, 

U=iPA. 

But from Art. 36 and 47 we have 


Therefore 


Pl = sl 
4 c 


1 P7 3 

and 


U 6 c 2 E 


( 222 ) 


But I = ak 2 (Art. 163) in which k is the radius of gyration of the 
cross-sectional area with respect to the neutral axis. Hence 


rr 1 fc 2 s 2 7 
U ~6 c 2 £7° 


or 


cV 6 (UE 
k \ al 


(223) 


If U is expressed in inch-pounds, E in pounds per square inch, a. 
in square inches, and k, c, and l in inches, s will be expressed in 
pounds per square inch. 

k 2 

Since the numerical value of the ratio is always the same 

for similar shaped sections, the amount of energy that a horizontal 
rectangular beam will absorb when the short dimension is vertical 
will be the same as that absorbed when the long dimension is* 
vertical. 













290 


IMPACT AND ENERGY LOADS 


PROBLEMS 


228. Show that if the beam has a rectangular cross-section equation 
(223) reduces to 

t/ = —= -)aZ,. (224) 

18 E 9 \2 El ’ 


which shows that the energy load which this beam can resist is only one- 
ninth as large as the energy load it could resist if it were used as a tension 
member and stressed to the same maximum value s. 

229. Show that if the beam has a circular cross-section, equation (223) 
reduces to 


U = —-al=— -\l. 

24 E 12 \2 EJ 


(225) 



Simple Beam with Uniformly Distributed Load .—The relation 
between a gradually applied load-and the deflection would be 

similar to that shown 
in Fig. 232(6). The 
final values of the load 
and maximum deflec¬ 
tion are W and A, 
respectively. The de¬ 
flection at any section 
is y and the load that 
aGts through this dis¬ 
tance is wdx (Fig. 233). 
Hence the energy load 
U that can be applied (which is equal to the work done on the 
beam) is 

U= f (iwdx-y), 

Jo 

and by making use of the expression for y found in Art. 46, we have, 

tt 1 P , W (IH h? , 

U ~2'J g wdx m\W “12+24 


Fig. 233.—Energy resistance of beam; uniform 
load. 


Hence 


_ 1 w 2 /l 5 l 5 l 5 \ 
— 2 ^/\48~48 ' 120 * 


U = 


1 w 2 l 5 
240 ^El 


(226) 










































EFFECT OF FORM ON ENERGY RESISTANCE 


291 


But w may be expressed in terms of the maximum unit-stress 
in the beam by means of the flexure formula, Art. 34; namely, 

\wl 2 = —. Therefore 
o o 




2J5 

El 


hence 


= i 1 S 1 

15c 2 JSf 


TT _ 4 /c 2 s 2 

17 15 c 2 E°^' 


(227) 


PROBLEMS 


230. Show that if the beam has a rectangular cross-sectional area equa¬ 
tion (227) reduces to 

4 


TJ - 


45 E 


-al 


-16 3 * 


(228) 


which shows that the energy load which this beam can resist is only as 
large as the energy load it could resist if it were used as a tension member 
and subjected to the same maximum unit-stress. 


231. Show that for a beam having a circular cross-sectional area equation 
227 reduces to 



al. 


(229) 


121. Effect of Form on Energy Resistance of Beams. —Equa¬ 
tions (223) and (227) and the equations given in Problems 228 
to 231 show that the material in a beam having a constant 
cross-section is inefficient in absorbing energy. For example, the 
expression in Problem 228 shows that a rectangular beam, when 
loaded at the mid-span with a concentrated load, can absorb only 
one-ninth as much energy as the same beam could absorb if all the 
material in the beam were stressed to the same degree. 

The inefficient use of material in a beam having a constant 
cross-section for resisting an energy load arises from two causes: 
(1) since the bending moment is relatively small at sections near 
the supports (for simple beams) the unit-stress, even in the outer 
fibers at these sections, is necessarily small, and hence the material 
toward the ends of simple beams can absorb very little energy; and 






292 


IMPACT AND ENERGY LOADS 


(2) the material near the neutral plane throughout the length of 
the whole beam can develop only small stresses and hence can 

/I s 2 \ 

absorb only a small amount of energy per unit volume (o^)* 


The first of these difficulties is largely overcome in leaf springs 
by making the leaves of gradually decreasing length (Fig. 234) 
so that the moment of inertia of the cross-sections decreases towards 
the supports approximately in proportion to the bending moment. 
Incidentally it is well to note that a leaf spring absorbs a consider¬ 
able amount of energy due to friction between the leaves. 




The second of the above difficulties may be overcome partly 
by placing as much of the material as far from the neutral surface 
as practicable, as for example, in the form of an I-section. If now 
a simple beam with an I-section is also made so that the cross- 
sectional area decreases towards the supports, as is sometimes done 
in forged axles, etc. (Fig. 235), the resistance of the beam to energy 
loads is very much greater than that of a beam having a constant 
cross-section that would have the same static strength as the beam 
with the variable cross-section. 

It is clear, therefore, that when a beam is subjected to an energy 
load the distribution of the stress in the beam should be as uniform 
as possible throughout the beam so that the energies absorbed by 


all unit volumes 



will be approximately equal. 


Again, the energy absorbed by a beam as shown by equations 
(223) and (227) increases with the volume of the beam provided 
the other influencing factors discussed above are constant. Thus, 
if a simple beam has a constant cross-section the amount of energy 
it will absorb increases as the length of the span of the beam is 
increased, whereas the static strength decreases with an increase 
in the length of the span. 







SPECIAL CASES OF ENERGY LOADS 


293 


122. Special Cases of Energy Loads on Beams. —Energy 
Load Due to a Falling Weight .—It is convenient frequently to 
express the unit-stress caused by an energy load, due to a falling 
weight W, in terms of the unit-stress that would be caused by the 
force W when gradually applied. Similarly, the deflection caused 
by a falling weight W may be expressed in terms of the deflection 
that would be caused by a load W when gradually applied. This 
may be done by substituting in equation (223) or (227) the value 
of U for the beam in question similar to the method used in Art. 
118; or it may be done as follows: 

If a body having a weight W falls from a height h on a simply 
supported beam the maximum deflection, A, of the beam will be 
proportional to the maximum unit stress, s, developed and the 
ratio of s to A is assumed to be the same, within the proportional 
limit, as that of the unit-stress, si, to the deflection, Ai, caused 
by a static load equal to W, thus, 


A___s 
Ai ~ sd 


(230) 


provided the proportional limit of the material is not exceeded. 

Further, if Q is a static load that causes a deflection, A, equal 
to that caused by the energy load, the work done by Q, which is 
\QA, will be equal to the energy supplied or given up by the 
falling body, and since the assumption is here made that all the 
energy of the falling body, W(h+A), is absorbed in stressing the 
beam, then JQA will be equal to the energy absorbed by the beam. 

Hpti pp 

W(h+A)=iQA .(231) 


But the static loads are proportional to the stresses they develop 
and hence from (1) we have 


A _ s __Q 
Ai si W’ 


(232) 


By combining (231) and (232) there is found 

s = si+si<^|l+^ and A = Ai+Ai-^l+— . . (233) 


in which s and A are the unit-stress and deflection, respectively, 
due to the falling body, and si and Ai are the stress and deflec- 






294 


IMPACT AND ENERGY LOADS 


tion, respectively, due to a static load equal to the weight of the 
falling body. Expressions for si and Ai are given in Chapters 
V and VI. Equations (233) show that a body which causes a 
relatively small stress and deflection when applied as a static 
load may cause a large stress and deflection if allowed to drop on 
the beam through a relatively short distance. 

Sudden Load. —If the value of h in equation (233) is zero, that 
is, if the load W is a sudden load (see Art. 118 for definition of 
sudden load), then the values of s and A due to the sudden load 
are 

s = 2si and A = 2Ai.(234) 

Therefore, a sudden load applied to a beam will cause twice the 
stress and twice the deflection that will be caused b}^ the same 
load when gradually applied. 

123. Deflection Due to Any Energy Load.—If an impact 

load U is delivered to a beam by any means other than that of a 
falling weight the deflection may be found from equation (233) as 
follows: After estimating the value of the energy U delivered to 
the beam, a value for W may be selected arbitrarily and a value of 
h that would be required to cause this weight W to deliver to the 
beam an amount of energy equal to U may be found. The unit- 
stress si and the deflection Ai that this weight would develop if 
applied gradually may also be found. Thus with values of h and 
Ai known, equation (233) may be used to find the value of A. A 
value of s in equation (233) may also be found by a similar method, 
but equations (223), (227), etc., are preferable for this purpose. 

ILLUSTRATIVE PROBLEM 

Problem 232. The proportional limit for hickory may be taken at 3000 
lb. per sq. in., and its modulus of elasticity at 1,500,000 lb. per sq. in. (a) 
Will a weight of 20 lb. falling 6 in. on the center of the span of a hickory 
beam 4 in. square cause a unit-stress above the proportional limit, if the beam 
has a span of 3 ft.? (Assume that the supports of the beam are rigid and that 
all the energy delivered by the falling weight is absorbed by the beam.) (6) 
How many inches will the beam deflect? (c) What should be the length 
of the span to make the stress equal 3000 lb. per sq. in.? 

Solution. —(a) First Method. —The deflection of the beam will be neglected 
in comparison with 6 in. and hence the energy U delivered to the beam is 


*7 = ^ = 20X6 = 120 in .-lb. 



ILLUSTRATIVE PROBLEM 


295 


The unit-stress caused by this energy load is (see Prob. 228). 

18 UE l l8X 120X1,500,000 
al \ 4X4X36 



= 2370 lb. per sq. in. 

(a) Second Method .—If the deflection of the beam is not neglected the 
maximum unit-stress s developed in the beam is 


S=Si+S 



The unit-stress Si due to a central static load of 20 lb. is 


Me 1 c 1 2 

. = _ 7 “ = “PL -= - X20X36X— 77 - 
7 4/4 *(4 )‘ 


= 16.9 lb. per sq. in. 


The maximum deflection, Ai, of the beam caused by a central static load of 
20 lb. is 

1 PI 3 1 20 X (36) 3 

1 ~48 E/~48 1,500,000X^(4) 4 

= 0.000608 in. 

Hence, 

I 2X6 

5 = 16.9 + 16.9 X l+ „ ~ 

\ 0.000608 

= 16.9+2365 
= 2382 lb. per sq. in., 

and hence the error introduced in the first method of solution by neglecting 
the deflection of the beam is very small. 


( 6 ) The maximum deflection of the beam is 

I 2 h 

-A 1 +A 1 yjl+- 

= 0.000608+0.000608 X140 
= 0.000608+0.0853 
= 0.0859 in. 


(c) Since the deflection may be neglected in determining the unit-stress, 
we have 

/l8 UE 

s = J —— = 3000. 

\ al 













296 


IMPACT AND ENERGY LOADS 


Hence 

18 UE 18X120X1,500,000 

^ ~ a X (3000) 2 _ 16 X (3000) I 2 

= 22.5 in. 


PROBLEMS 


233. Three beams, A, B and C, all of the same dimensions are arranged 
as shown in Fig. 236 similar to one type of draft gear. Each beam is 3 in. 

square and has a span of 12 in. (a) 
What static load applied at the mid-span 
will cause a maximum unit-stress of 

- ^ 20,000 lb. per sq. in.? (6) Will the beams 

I I r-i when arranged as shown in Fig. 236 

-wL resist a greater static load than any one 

* of the beams? (c) What is the weight 

Fig. 236. 0 f a body which, when allowed to fall 

a height h of 8 in. will cause a stress 
of 20,000 lb. per sq. in? ( d ) What stress would the weight in (c) cause if it 
were resisted by only one of the three beams? Ans. (a) 30,000 lb.; (5) no; 
(c) 30 lb.; (d) 34,500 lb. per sq. in. 


¥ 


234. A steel bar 2 in. in diameter and 3 ft. long remains in a horizontal 
position as it falls a distance of 4 ft. and strikes rigid supports at its ends. 
If all the energy is absorbed by the beam what unit-stress is developed in the 
beam, assuming that the proportional limit is not exceeded. 


235. A high carbon steel cylindrical shaft has a diameter of 4 in. It is 
used as a simple beam with a span of 6 ft. The proportional limit of the mate¬ 
rial is 60,000 lb. per sq. in. If a moving body strikes the beam at mid-span 
and two-thirds of the energy of the moving body is absorbed by the beam, how 
much energy does the body possess if it produces a unit-stress equal to one- 
half of the proportional limit? Ans. 1690 in-lb. 









CHAPTER XIV 


REPEATED LOADS. FATIGUE OF METALS 

124. Introduction.—A repeated load is a force that is applied 
many times to a member, causing stress in the material that con¬ 
tinually varies, usually through some definite range. If a stress 
is developed in a member and is then released the member is said 
to have been subjected to a cycle of stress. Further, if a tensile 
stress has been developed and when released a compressive stress 
is developed and this stress is then released, the member is said 
to have been subjected to a reversed cycle of stress or, briefly, to a 
reversal of stress; the reversal of stress is said to be complete if the 
opposite stresses are of equal magnitudes. For example the piston 
rod of a steam engine that runs at 300 r.p.m. 10 hours per day 
300 days per year for 10 years is subjected to 540,000,000 cycles 
of approximately complete reversals of axial stress; a car axle is 
subjected to about 50,000,000 complete reversals of bending stress 
in its normal “life”; a band saw in a normal service of about 
two months is subjected to about 10,000,000 cycles of stress, 
the stress in each cycle ranging from approximately zero to a 
maximum, etc. 

Experience and experiments have shown clearly that the 
resistance of rolled or forged iron and steel (wrought ferrous metals) 
to repeated loads depends on very different action in the material 
than does the resistance to static or impact loading. For example, 
iron and steel will rupture when subjected to millions of reversals 
of stress not only when the calculated unit-stress in the material 
is less than the static ultimate strength of the material but even 
when the calculated unit-stress is less than the static proportional 
limit. 

The failure of a material caused by repeated loads is a gradual 
or progressive failure. The failure seems to start at some point in 
the material at which the stress is much larger than the calculated 
stress, and this high localized stress develops a small crack which 
gradually spreads, as the load is repeated, until the whole member 

297 


298 


REPEATED LOADS. FATIGUE OF METALS 


fails. The apparent loss of strength of a material due to repeated 
stress is frequently called fatigue) thus a fatigue failure is a failure 
caused by repeated loads. Failures of various machine members 
due to repeated stress have frequently occurred in service: Rail¬ 
way axles, steam turbine shafts and discs, crank shafts, piston 
rods, valve rods, springs in automobiles, etc., give much trouble in 
this respect. 

Experimental Investigations .—In 1870 Wohler published 1 the 
results of an extensive series of tests of various grades of iron and 
steel subjected to repeated direct tensile and compressive loads, 
to repeated bending loads, and to repeated torsional loads. The 
tests were carried out for the Prussian Railways during a period 
of about ten years. They are considered a classic in this field 
of investigation and until recently were the main source of our 
knowledge of resistance of material to repeated loads. 

The number of cycles of stress to which the material was sub¬ 
jected in Wohler’s tests was usually less than 1,000,000, although 
a small number of specimens were stressed 10,000,000 times, a 
very few specimens were stressed 40,000,000 times, and one spe¬ 
cimen 132,000,000 times. Although Wohler’s experiments gave 
reliable information for the machines and materials used at that 
time, the development of high speed machinery and of alloy and 
heat-treated steels have created a need for further experimental 
investigations. In recent years several important investigations 2 
have been made that have added much to our knowledge of the 
subject. 

125. Endurance Limit. —The endurance limit of a material 
is the maximum unit-stress that can be repeated, through a definite 
cycle or range of stress, an indefinitely large number of times 
without causing the material to rupture. As will be discussed 
later, the larger the cycle or range of stress is made the smaller the 
value of the endurance limit becomes; but when the term endur¬ 
ance limit is used without any limiting' statement as to range of 
stress it will be understood to be the endurance limit with com¬ 
pletely reversed cycles of stress. 

1 A summary of Wohler’s work is given in English in “ The Testing of 
Materials of Construction ” by Unwin. 

2 See Bulletins 124, 136 and 142 of the Engineering Experiment Station 
of the University of Illinois; a bibliography is given in Bulletin 124 and ref¬ 
erences to later work is given in Bulletins 136 and 142. 


ENDURANCE LIMIT 


299 


S-N Diagrams .—If several specimens are cut from the same bar 
of forged or of rolled steel 3 and are subjected to repeated complete 
reversals of stress (see Art. 156 for description of repeated-stress 
testing machines), it will be found that when a specimen is stressed 
nearly to the ultimate strength of the material in each cycle of 
stress the specimen will rupture after being subjected to a small 
number of cycles of stress; if a second specimen is tested in the 
same way but stressed slightly less than the first, a larger number 
of reversals of stress will be required to cause the specimen to 
rupture. Now, if a series of such experiments are carried out, 
the maximum unit-stress in any specimen being somewhat less 
than in the preceding specimen, the relation between the value of 
the completely reversed unit-stress, s, and the number of reversals y 
N, required to rupture the specimen will be found to be repre¬ 
sented by a curve similar to that shown in Fig. 237 in which stresses 
are plotted as ordinates and numbers of reversals as abscissas. 

The curve in Fig. 237 is 
called an s-N curve, and the 
ordinate to the s-N curve 
where the curve has become 
approximately horizontal is 
taken as a measure of the 
endurance limit of the ma¬ 
terial as defined above. Thus 
the endurance limit ( s r ) ob¬ 
tained from the curve in 
Fig. 237 is approximately 
±19,000 lb. per sq. in. 

That is, this material will rupture when subjected, in bending, 
to several million cycles of completely reversed stress if the max¬ 
imum unit-stress in each cycle is slightly greater than 19,000 lb. 
per sq. in. 

Another way of obtaining an s—N curve and the endurance 
limit is to plot values of the logarithms of s and N (or the equiva¬ 
lent of this, namely, to plot values of s and N on logarithmic paper). 
When this is done the s-N curve (Fig. 238) is a straight sloping 
line until it changes its slope rather abruptly and becomes hori- 

3 Sufficient experimental data for castings of iron or steel and for non- 
ferrous metals from which to draw definite conclusions are not yet available. 



Reversals of Stress Causing Failure, 
in Millions (N) 

Fig. 237.— An s-N diagram for steel 
subjected to completely reversed 
cycles of bending stress. 



















300 


REPEATED LOADS. FATIGUE OF METALS 


zontal. The unit-stress at which the change of slope occurs is 
taken as the measure of the endurance limit ( s r ). 

Another convenient method of obtaining an s-N curve and the 
endurance limit is to plot the unit-stresses as ordinates and the 
logarithms of the numbers of reversals as abscissae, sometimes 
called semi-logarithmic plotting. The curves thus obtained are 
very much the same as those in Fig. 238. Thus, in Fig. 239, are 
shown s-N diagrams for several grades of steel plotted in two 
ways: In Fig. 239(a) values of s and N are plotted as Cartesian 
coordinates and in Fig. 239 (b) is shown the semi-logarithmic 
plotting of the same values. 

As shown in Fig. 238 and 239, wrought ferrous metals, if 
subjected to complete reversals of stress, will usually fail after 
resisting 1,000,000 to 5,000,000 cycles of stress, when the max¬ 
imum stress in each cycle is slightly above the endurance limit. 
(Compare these values with the probable number of repetitions 
of stress in the “ lifetime ” of various members as given in Table 6 
and determine if these members can safely be stressed above the 
endurance limit of the material.) 


TABLE 6 


Part of Structure or Machine 

Approximate num¬ 
ber of repetitions of 
stress in the “ life¬ 
time ” of the struc¬ 
ture or machine. 

Railroad bridge, chord members. 

Elevated railroad structure, floor beams. 

Railroad rail, locomotive wheel loads. 

Railroad rail, car wheel loads. 

Airplane engine, crankshaft. 

Car axles... 

Automobile engine, crankshaft. 

Line shafting in shops. 

2,000,000 

40,000,000 

500,000 

15,000,000 

18,000,000 

50,000,000 

120,000,000 

360,000,000 

1,000,000,000 

15,000,000,000 

250,000,000,000 

Steam engine, piston rods, connecting rods and crank¬ 
shafts . 

Steam-turbine shafts. 

Steam-turbine blades. 



126. Localized Stress and Fatigue Failure. —When a ductile 
steel specimen is caused to fail by a gradually increasing (static) 

















Unit Stress in lb. per sq. in. (s) 


LOCALIZED STRESS AND FATIGUE FAILURE 


301 



Fig. 238.—Logarithmic s-N diagrams for steel subjected to 

reversed cycles of bending stress. 


completely 



(a) (&) 

Fig. 239.— s-N diagrams for steel subjected to completely reversed cycles of 
bending stress; (a) Cartesian plotting; (6) semi-logarithmic plotting. 















































































































































































































































302 


REPEATED LOADS. FATIGUE OF METALS 


load there is visible evidence of structural damage (plastic defor¬ 
mation) and of the approaching of failure, considerably before the 
failure occurs. If a specimen of the same material, however, is 
caused to fail by a repeated load, there is no plastic deformation or 
other warning of the approaching of failure. Thus the fatigue 
failure of a ductile steel is similar to that of a static failure of a 
brittle material; the fractured area of a ductile steel specimen that 
is broken by repeated loads, usually presents a crystalline appear¬ 
ance similar to that of a coarse-grained brittle material, whereas 
the fractured area of the ductile steel specimen when broken by a 
static load presents a “ silky ” or “ fibrous ” appearance. 

These observations suggested the crystallization theory of failure 
of steel due to repeated stress. It was thought that, in service, 
the repeated application of a load changed the steel from a fibrous 
ductile material to a crystalline brittle material. It is now known, 
however, that iron and steel are always crystalline and that the 
crystalline theory is entirely erroneous. 

Localized Stress Theory. —The most satisfactory explanation of 
a fatigue failure is the localized stress theory , the main features 
of which may be explained as follows: 

In determining the relation between stresses and static loads 
as expressed by the equations developed in the preceding chapters 
it was assumed 

1 . That the material was homogeneous; that is, there existed 

no discontinuities in the material and no abrupt changes 
in the properties of the material throughout the body. 

2. That there was a definite regularity of stress distribution 

on any section of the member: that is, no discontinuities 
or abrupt changes occurred in the distribution of the stress 
over the section. 

It is known, however, that these conditions never exist in struc¬ 
tural members; metals for example are composed of crystalline 
grains whose strength and stiffness vary, and there are local con¬ 
centrations of stress (localized stresses) at various portions of a 
member that may be much larger than the calculated values based 
on the assumed regularity of distribution. These localized stresses 
and irregularities in the stress distribution are due (1) to discon¬ 
tinuities in the material itself such is small flaws, fissures, non- 
metallic inclusions, etc., at the edges of which high stress exist 
and discontinuities in the properties of the material due to the 


LOCALIZED STRESS AND FATIGUE FAILURE 


303 


variation in the strength and stiffness of the crystalline grains 
from those assumed for the material as a whole. Both of these 
causes of localized stress will be referred to as internal disconti¬ 
nuities, and (2) to external discontinuities of the material such as 
abrupt changes of sections, etc., where the stress distribution is, 
as a rule, radically different from that assumed. 

Now when a ductile material is subjected to a static load, 
localized stresses that are considerably greater than the yield- 
point of the material may be developed (in fact the localized 
portion may even rupture) without seriously affecting the strength 
or deformation of the member as a whole, and hence the above 
assumptions are, in general, justified when the load is a static 
load (or is applied only a few times). But when the load is 
applied a very large number of times these localized stresses have a 
determining effect on the strength of the member (see Appendix 
III). 

It seems clear that if a steel specimen ruptures when subjected 
to a repeated load, the stress in the material must have reached 
its ultimate strength even though the calculated stress may be less 
than the static proportional limit of the specimen. The rupture 
starts at a point of high localized stress or at a point where a weak 
crystal occurs, and gradually spreads until the whole member 
ruptures. The first experimental evidence of the gradual spread 
of the area of rupture was obtained by Ewing, Humphrey and 
Rosenhain. 4 They found that when the localized stress in cer¬ 
tain crystals becomes sufficiently great the crystals yield by micro¬ 
scopic movement or sliding along their cleavage planes: These 
planes are called slip planes and their traces on a polished section 
of the member are seen under the microscope as dark lines, called 
slip lines or slip bands (see Fig. 240). For example, slip lines 
were detected in some of the crystals of a Swedish iron specimen 
when subjected to a few complete reversals of stress of ±20,000 
lb. per sq. in., although the static yield-point and ultimate strength 
of the material were 31,600 and 52,800 lb. per sq. in., respectively; 
after more rev( r ,als of stress were applied additional slip lines 
appeared and the original ones broadened. Finally, various 
groups of slip lines united forming a visible crack which gradually 
extended until rupture occurred. (See Fig. 2406.) 

4 Philosophic Transactions Royal Society A, Vol. 200, p. 241, 1903. 


304 


REPEATED LOADS. FATIGUE OF METALS 


However, not all slip lines develop into cracks, and there may 
be a development of slip lines and, possibly, of small cracks without 
resulting in failure of the member. In fact, fatigue failures fre¬ 
quently are not due to the development of slip bands but are 
caused by direct tearing apart at the points of high localized stress 
with subsequent spreading of the area of rupture, resulting in a 
failure that gives no warning of its approach. But, in any case, 
the cause of the failure is the localized stress due to internal or 
external discontinuities which are neglected in the usual formulas 
in mechanics of material as developed in the preceding chapters. 



Fig. 240.—Photomicrographs of steel: (a) View before stressing, (6) view 
after application of several thousand reversals of large stress showing 
slip bands and crack. (Obtained by Prof. H. F. Moore.) 


For example, in Fig. 241, is shown the section of a ruptured 
bolt that failed after being subjected to many repetitions of a ten¬ 
sile load. The failure started at the root of the thread where 
high localized stress occurred and spread inwardly. The dark 
portions of the cross-section show the area over which the crack 
spread and the light portion shows the area of rupture over which 
the material gave way suddenly. 

127. Values of Endurance Limits with Completely Reversed 
Bending Stress. —Values of the endurance limits of several grades 
of steel when subjected to completely reversed bending stress are 
given in Table 7. Values of the static tensile proportional limits 
and of the tensile ultimate strengths of the steel are also given in 
the same table. 



VALUES OF ENDURANCE LIMITS 


305 


The values in Table 7 show that the endurance limit for com¬ 
plete reversals of stress may be considerably below the static 
proportional limit of the material. In other words, the ultimate 
resistance of steel to repeated loads is often considerably less than 
the static elastic resistance (proportional limit) of the material. 
Further, if a heat treatment greatly increases the static propor¬ 
tional limit it does not necessarily follow that the endurance limit 
is raised proportionally. In fact, tests show clearly that the endur- 



Fig. 241.—View of area of rupture of bolt that failed due to repeated stress, 
showing evidence of progressive failure. (Obtained by Prof. H. F. 
Moore). 

ance limit of steel bears a much more consant relation to the 
ultimate strength than to the elastic limit or proportional limit, 
and that tor most rolled or forged steels the endurance limit with 
completely reversed bending stress is approximately 0.45 of the static 
tensile ultimate strength (s r = 0A5s u ) as maybe seen by referring to 
Table 7. 

128. Relation of Endurance Limit with Direct Axial Stress to 
Endurance Limit with Bending Stress. —The endurance limit, s d , 
of a material when subjected to complete reversals of axial stress 


Values of Endurance Limit, Ultimate Strength, Proportional Limit, and of the Ratio of the Endurance Limit to 
the Ultimate Strength and to the Proportional Limit for Various Grades of Steel 
(Specimens were subjected to completely reversed cycles of bending stress and were free from abrupt changes of section or surface scratches.) 


•306 


REPEATED LOADS. FATIGUE OF METALS 


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RELATION OF ENDURANCE LIMITS 


307 


(stress varying from a direct tensile to an equal direct compression 
stress) is found to be lower than the endurance limit, s r , of the 
same material when subjected to complete reversals of bending 
stress. 

A safe approximate value of s d for rolled or forged iron and steel 
is 

s d = 0.Qs r .(235) 

The reason why s d is less than s r is due probably to the fact that 
the load always is applied with some eccentricity and hence the 
stress in the specimen is greater than the calculated stress; and 
also to the fact that in the bending-stress specimen only the outer 
fibers are subjected to the maximum stress whereas in the axial- 
stress specimen all the fibers are subjected to the same nominal 
stress. 

129. Relation of Endurance Limits in Torsion and Bending.— 

The resistance of material to repeated shearing stress is of impor¬ 
tance in some machine members such as torsional springs, shafts, 
etc., and although the number of repeated stress tests with reversed 
torsion is much less than with reversed bending, the tests are rea¬ 
sonably consistent in showing that the endurance limit of steel 
with complete reversals of torsional shearing stress, denoted by 
(s r )„, is 0.55 time the endurance limit with complete reversals of 
bending stress. Thus, 

(s r ) s — 0.55$ r .(236) 

130. Localized Stress Due to External Discontinuities.—(a) 

Abrupt Change of Section .—The results of several independent- 
experimental investigations 5 have shown that high localized 
stresses occur at abrupt changes of section of a member when the 
member is resisting a load. Consequently, a repeated stress 
specimen having an abrupt change of section such as a square 
corner, keyway, screw thread, groove, etc., is found to fail when 
the calculated stress (found by use of the ordinary equations of 
mechanics of materials) is lower than the endurance limit of the 
material as found from specimens of the same material that are 
free from abrupt changes of sections. The ratio, then, of the 
endurance limit of the material to the endurance limit of the spe¬ 
cimen with the abrupt change of section is a measure of the localized 

5 See Appendix III. 




308 


REPEATED LOADS. FATIGUE OF METALS 


stress due to the abrupt change of section. For example, if the 
endurance limit of specimens free from abrupt changes of section 
is found to be 30,000 lb. per sq. in. and the endurance limit of 
specimens having given abrupt changes of section is found to be 
15,000 lb. per sq. in., then the actual (localized) stress in the latter 
specimens is twice the calculated stress. Thus, repeated stress 
tests may be used to determine the value of localized stress. (Other 
methods of determining localized stresses are discussed in Appen¬ 
dix III.) 

TABLE 8 


The Reduction in the Endurance Limit of a Specimen of Steel 
Subjected to Reversed Bending Due to Abrupt Change in 
Section; Specimens were 0.4 Inch in Diameter. 


Form of Section 

Reduction in Endur¬ 
ance Limit in Per 
Cent 

Groove with 10-in. radius. 

0 

Groove with 1-in radius. 

5 

Groove with j-im radius.. . 

10 

Shoulder or notch with small fillet. 

25 

Square shoulder. 

50 

Sharp V-notch. 

65 


It is very clear from the results in Table 7 that, in designing 
members that are subjected to repeated loads, sharp corners should 
be avoided as much as possible, and experience has emphasized 
the truth of this statement. For example, the gears, crank shafts, 
and connecting rods of the Liberty airplane engine, at first, fre¬ 
quently failed from repeated stress; the failure in the gears started 
at the sharp corner at the bottom of the teeth, of the crank shafts 
at the corner of the keyways, and of the connecting rods at the 
sharp corner where the connecting rod bolt-head fitted on the 
assembly. Further, the British found that, when a failure of an 
airplane crankshaft was due to a sharp corner, failure after failure 
would occur in engines of the same type in approximately the same 
number of running hours. The trouble in the Liberty motor parts 
finally was practically eliminated in all parts by adopting a small 
fillet of about in. radius at each corner. 













WORKING STRESS WITH REPEATED LOADS 


309 


(b) Surface Finish— Experiments 6 have shown clearly that 
the surface finish of a member has an appreciable effect on its 
endurance limit, the smoother finish giving the higher value for 
the endurance limit. Since scratches, tool marks, indentations 
due to cold pressing, etc., may be regarded as more or less abrupt 
changes of section, giving rise to localized stress, this effect of 
surface finish is to be expected. 

The results of experiments 6 show that a specimen having a 
surface produced b}^ turning a finishing cut in a lathe may have an 
endurance limit as much as 10 per cent less than a specimen having 
a surface produced by fine grinding, and somewhat rougher turn¬ 
ing may cause a reduction in the endurance limit of 20 per cent or 
more. Although fine grinding does not produce as good results as 
does polishing, it would probably be considered satisfactory for 
many commercial machine members. 

It is clear, therefore, that surface scratches due, for example, 
to grit in bearings, indentations due to hammer blows or high 
bearing pressure, tool and die marks in cold pressed work, etc., 
must be taken into account in estimating the resistance of a 
member to repeated loads. 

For example, repeated stress failures of the cold pressed water 
jackets on certain airplane engines have been found to be due to 
die marks. And, the observation of fractures that have occurred 
in service shows that slight surface scratches appear to determine 
the position of a fracture in steel subjected to reversals of stress. 

131. Working Stress with Repeated Loads.—The endurance 
limit of steel subjected to millions of cycles of stress is really the 
fatigue ultimate strength of the material since if the unit-stress 
developed in the material is greater than the endurance limit the 
material will rupture. Further, there is no unit-stress below the 
endurance limit at which the beginning of inelastic action or struc¬ 
tural damage can be detected, whereas with static loading the 
beginning of structural damage can be detected at stresses much 
below the static ultimate strength of the material since the pro¬ 
portional limit, yield point, etc., indicate the beginning of inelastic 
action or structural damage caused by static loads. It seems clear, 
therefore, that the working stress for steel members subjected to re¬ 
peated stress should be based on the endurance limit of the material. 

If, however, the endurance limit of the steel is not known 
6 Bulletin 124, Engineering Experiment Station, University of Illinois. 


310 


REPEATED LOADS. FATIGUE OF METALS 


(see Art. 156 for machines and methods of testing) an approximate 
value of the endurance limit, may be found from the static ulti¬ 
mate strength of the steel since tests have shown, as stated in 
Art. 127, that the endurance limit for completely reversed cycles 
of stress may be taken with reasonable accuracy as 0.45 (or roughly 
0.5) of the static tensile ultimate strength of the material. 

Further, if the stress is not completely reversed, an approxi¬ 
mate value of the endurance limit for a given range of stress may 
be obtained from the endurance limit with completely reversed 
stress by means of equation (240) or (241) in Art. 132. 

Now it is important to note that the values of the endurance 
limits found from the usual laboratory tests are obtained with a 
specimen having a very gradual change in section and having a 
polished surface finish, and hence the endurance limit of the 
specimen may be considered to be the endurance limit of the 
material, the internal discontinuities being assumed constant for a 
given material. Again, it is only when these conditions exist 
that the endurance limit may be assumed to be 0.45 (or roughly 
0.5) of the static tensile ultimate strength of the material. 

Need for Margin of Safety .—As discussed in Art. 7, three of the 
reasons for selecting a working stress less than the maximum 
usable strength (yield point, etc.) of the material, in the case of 
static loading, are (1) that the actual loads to which the member 
will be subjected are seldom known with certainty, (2)that the actual 
stresses in a member, even if the member were subjected to known 
loads, may be considerably greater than those calculated by the 
usual formulas (as developed in the preceding chapters), since 
these formulas do not take account of the localized stresses due to 
various causes (see Art. 138 and 143) and (3) that the properties 
of the material vary, due to variation in the quality of the material, 
from those obtained from the tests of sample specimens. 

Now the uncertainty of the loads in the case of repeated loading 
is probably about the same as in static loading, and about the 
same allowance in selecting the working stress should be made for 
the uncertainty of the load with both types of loading. Further, 
the uncertainties in the calculated stresses in a member due to the 
presence of localized stress are probably no greater with repeated 
loading than with static loading, but the effect of localized stresses 
have a very much greater effect on the resistance of the material to 
repeated loads than on the resistance to static loads. 


WORKING STRESS WITH REPEATED LOADS 


311 


For, as discussed in Art. 126 and 143, and in Appendix III, when 
localized stresses that are considerably in excess of „ the calcu¬ 
lated stresses occur in a steel member that is subjected to static 
loads (and such localized stresses always exist) the material at the 
point where a localized stress occurs will yield, since the steel is 
fairly ductile, and the stress at this point will be relieved, thereby 
transmitting some of the excess stress to the surrounding, less- 
stressed material, and the resistance of the member as a whole 
will not be seriously affected by this action; whereas, if the same 
member were subjected to repealed loads, experience and experi¬ 
ments show that a minute crack would probably start at the point 
of high localized stress and would gradually spread, as the loads 
were repeated, until the whole member would rupture. 

Now, since localized stresses are so important in repeated 
stress members and since experimental data are now available by 
means of which a reasonably close estimate of the intensity of 
the localized stress in a member can be made, when the localized 
stress is due to external discontinuities such as abrupt changes of 
section, surface finish, etc., it seems unnecessarily indefinite 
to make allowance for localized stress in repeated-stress members 
by selecting, arbitrarily, low working stresses. 

Method of Calculating Maximum Stress. —The method of obtain¬ 
ing the maximum (localized) unit-stress due to external discon¬ 
tinuities is as follows: Calculate, according to the usual formulas 
used for static loading, the maximum unit-stress that the loads 
cause, and then multiply this unit-stress by a factor to obtain the 
intensity of the localized stress, the factor depending on the form 
of the member. The value of the factor is obtained from the 
results of repeated-stress experiments and from other methods 
discussed in Appendix III. Approximate values for these factors 
for steel members having various external discontinuities are given 
in Table 9. The value of the factor is unity if the member has a 
very gradual change of section and a very smooth surface, that is, 
if the member has no external discontinuities. 

Selection of Working Stress. —Now, as explained in Art. 126, 
localized stresses arise from internal discontinuities as well as 
from external discontinuities, and the localized stresses due to the 
internal discontinuities, such as the stresses that occur at the edges 
of the minute blowholes, pipes, etc., and at points of high-bearing 
stress of two or more crystals due to a disadvantageous arrange- 


312 


REPEATED LOADS. FATIGUE OF METALS 


ment of the crystals, etc., cause a specimen even when free from 
external discontinuities to rupture, when subjected to repeated 
stress, at a calculated unit-stress less than the static elastic limit 
of the material. In other words, the unit-stress on some minute 
area in the specimen exceeds the ultimate strength of the material 
at that point even when the maximum calculated stress in the 
material is considerably less than the static elastic limit of the 
material. 

TABLE 9 

Maximum Localized Stresses Due to Various External Discontinuities 

Concentric Groove around 
Cylindrical Shaft: 

Ratio of maximum localized unit- 


Ratio of radius of groove or fillet to stress to computed max imum 

diameter of shaft. unit-stress in shaft. 

0.1 . 2.0 

0.5.. 1.6 

1.0 . 1.2 

2.0 . 1.1 

Special Types of Abrupt Changes of 
Section and of Surface Finish: 

Square corner. 2.0 

Sharp V-thread. 3.0 

Whitworth thread. 2.0 

U. S. Standard thread.. 2.5 

Surface finish produced by lathe tool; 

(Scratches due to grit in bearings). 1.2 

Surface finish produced by a grinding wheel... 1.05 


But, there is no reliable way of detecting or of measuring the 
effects of internal discontinuities as can be done for the external 
discontinuities. Therefore, an allowance is made for the effect of 
these internal conditions by selecting a working stress as some 
proportion of the endurance limit found from the usual repeated- 
stress tests. Experiments indicate that specimens taken from 
the same bar of steel and made and tested as nearly alike as pos¬ 
sible rupture, when subjected to repeated loads, at calculated unit- 
stresses that may vary considerably, and the endurance limits of 
specimens taken from different bars of a shipment of steel of pre¬ 
sumably the same quality may vary greatly. 

Thus, even when the limiting values of the repeated loads are 
known (as for example in the case of a spring provided with 











ILLUSTRATIVE PROBLEMS 


313 


tops to prevent excessive deflection), and even when due allow¬ 
ance has been made, as discussed above, for the effect of the 
external discontinuities of the member, it is recommended that the 
working stress should not be greater than one-half the endurance 
limit in order to make allowance for the effect of internal dis¬ 
continuities, and that the working stress should not be greater 
than one-third of the endurance limit in order to make allowance 
for the uncertainties in the assumed loads, in addition to the uncer¬ 
tainties in the internal conditions of the metal. However, in using 
a working stress equal to one-third of the endurance limit it is 
assumed that the maximum calculated stress in the material is the 
localized stress arising from eternal discontinuities as discussed 
above, and not merely a nominal stress found from the ordinary 
equations of mechanics of materials. 

It is assumed throughout this chapter that the repeated loads 
are not applied with impact. If a member is subjected to repeated 
impact loads some additional allowance must be made for the effect 
of the impact, depending on the conditions of the problem. 

Again, some machine members in service are subjected to 
repeated stresses greater than the endurance limit and are not 
expected to resist these stresses indefinitely as for example, band 
saws for wood, wire rope running on sheaves, ball bearings, gun 
barrels, etc. 

ILLUSTRATIVE PROBLEMS 


Problem 236. The crank pin (Fig. 242) has a diameter, d, of 5 in. and a 
length of 6.25 in., these dimensions being determined from the allowable 
bearing pressure. The shaft is made of the 
normalized 0.57 per cent carbon steel listed in 
Table 7. The radius of the fillet at section A A is 
^ in. The connecting rod exerts a total pressure 
of 25,000 lb. on the pin. If the pressure is assumed 
to be uniformly distributed, will the pin probably 
fail, due to repeated stress, in a service of ten 
years? If not, what is the ratio of the localized 
stress to the endurance limit? The speed of the 
shaft is 120 r.p.m., the engine runs ten hours per 
day and three hundred days per year. 

Solution .—The crank pin would be subjected to about 200,000,000 com¬ 
plete reversals of bending stress and hence would fail if the unit-stress in the 
pin exceeds the endurance limit of the material. The endurance limit of this 
material, as given in Tab’e 7, is 33,000 lb. per sq. in. 

















314 


REPEATED LOADS. FATIGUE OF METALS 


The maximum unit-stress in the pin is found as follows: The nominal unit- 
stress as given by the flexure formula is 


Me 25,000X3.125X2.5 

T~ 


= 6360 lb. per sq. in. 


and by making use of Tables 8 and 9 as a guide, the maximum unit-stress at 
the root of the fillet will be estimated to be twice the calculated value. Thus, 

maximum s = 6360X2 = 12,720 lb. per sq. in. 

which is less than the endurance limit. The ratio, /, of the localized stress to 
the endurance limit, then, is 

12,720 _ 1 
33,000“ 2.6’ 

and hence the stress in the crank pin is slightly greater than the permissible 
working stress recommended. 

Problem 237. A piston rod (Fig. 243) is to be subjected to a maximum load, 
P, of 60,000 lb. The engine runs at a speed of 150 r.p.m. and will be assumed 

to be in service six hours per day, 
three hundred days per year for 
at least ten years. The rod is 
made of 0.4 per cent carbon steel 
heat treated to give an endurance 
limit with completely reversed 
bending stress of 46,700 lb. per sq. 
in. (compare with Table 7). The 
rod has sharp corners at B, and 
the diameter of the rod at the 
roots of the threads may be 

assumed to be 0.85 d. What should be the diameter, d, of the rod? 

Solution .—The endurance limit for the material (see Art. 128) is, 
s d = 46,700X0.6 = 28,000 lb. per sq. in. 

If the working unit-stress ( s w ) is taken as one-third of the endurance limit, 
then, 

28,000 

s w = — -= 9330 lb. per sq. m. 


«f= 


n. In 


A d i*r< 7 - 

A. ,t *7 

14"rad. 

u 

111-Li 


Fig. 243. 


Now the average or nominal unit-stress on the cross-section at the roots of the 
thread is 

P 60,000 
Sav *“ a~ir(0.85d) 2 ’ 


and the maximum (localized) unit-stress at the sharp corner, B, or at the root 
of the threads, will be two or three times the average unit-stress. A value of 





















PROBLEMS 


315 


3 will be used since it is likely that the stress would be far from constant over 
the area due to eccentricity of loading even if there were no abrupt changes of 
section. Therefore, 


Hence 


s max. 



a 


9330 = 3 


60,000X4 

7r(0.85d) 2 


d = 5 .83 in. 


PROBLEMS 


238. The rods which operate the slide valves of engines frequently break- 
One method of construction is shown in Fig. 244. If the threads on the rod 
are U. S. standard threads, estimate a safe working stress for the rod if the 
material is hot-rolled low-carbon steel (about 0.20 per cent carbon, Table 7); 
also find the corresponding working value of the load P if the cross-sectional 
area of the bar is 2 sq. in. 



239. Fatigue failures frequently occur at the corners of cotter holes. If a 
bar having a rectangular cotter hole (Fig. 245) with square corners is subjected 
to repeated direct stress varying from zero to a maximum, and the bar is made 
of 0.40 per cent carbon steel having a tensile ultimate strength of 80,000 lb. 
per sq. in., what working unit-stress should be used in the design of the bar? 
What working axial load, P, should be applied to the bar if the cross-sectional 
area of the bar is 4 sq. in.? 

240. A circular shaft having a constant diameter of 2 in. is to be subjected 
to several million complete reversals of bending stress. The shaft is turned 
with a lathe tool. What is the maximum bending moment that should be 
applied to the shaft, if the shaft is made of about 0.50 carbon steel, hardened 
and tempered (see Table 7). 

241. In Fig. 246 are shown two forms of crank pins. The dimensions l 
and d of the crank pins are equal and the pins are made of the same grade of 


























316 


REPEATED LOADS. FATIGUE OF METALS 


steel. Compare the maximum repeated pressures the two pins can resist 
without breaking due to repeated complete reversals of bending stress. 



Fig. 246. 


132. Effect of Range of Stress.—The investigations of Wohler 
and the supplementary work of Bauschinger showed that the num¬ 
ber of cycles of stress required to cause rupture depends on the 
range of stress. For example, if the unit-stress in each cycle 
varies from a tensile stress of 30,000 lb. per sq. in. to a compres¬ 
sive stress of 30,000 lb. per sq. in. (a range of stress of 60,000 lb. 
per sq. in.) fewer repetitions are necessary to cause rupture than 
if the unit-stress in each cycle varies from zero to 30,000 lb. per 
sq. in. although the maximum unit-stress is the same for each of 
these ranges of stress. 

Further, the maximum or limiting unit-stress that can be 
repeated an indefinitely large number of times (the endurance 
limit) depends on the lower value of the stress in the cycle of stress 
and hence indirectly on the range of stress in the cycle. A formula 
based mainly on the results of Wohler’s experiments, that expresses 
the relation between the endurance limit of a material and the 
range of stress was developed independently by Goodman 7 and 
by J. B. Johnson, 8 and will here be referred to as the Goodman- 
Johnson formula. It may be derived as follows: 

Goodman-Johnson Formula .—In Fig. 247 the minimum unit- 
stresses (s mIn .) or lower limits of the various ranges of stress (A 5 ) 
which, in Wohler’s experiments, resulted in failure after the appli¬ 
cation of about 4,000,000 cycles of stress were plotted as ordinates, 
the abscissas being selected arbitrarily so that a straight line 
(. EOC ) would connect the ends of the ordinates. It was then 
found that if the corresponding maximum unit-stresses (s max ). or 
upper limits of the ranges of stress were plotted as ordinates the 
ends of the ordinates fell approximately on the straight line CAD 

7 Mechanics Applied to Engineering, p. 634. 

8 Johnson’s Material of Construction, 5th Edition, p. 781. 























EFFECT OF RANGE OF STRESS 


317 


(Fig. 247) where OA is equal to one-half of the static ultimate 
strength of the material (§ s u ), and MD and ME are equal to 
one-third of the static ultimate strength (Js M ). Thus, for com¬ 
plete reversals of stress the endurance limit was found by Wohler 
to be equal to about Js w , 9 and for a range of stress with the mini- 



Fig. 247.—Goodman-Johnson diagram. Effect of range of stress on endur¬ 
ance limit of steel. 

mum stress equal to zero the endurance limit was equal to about 

Now from Fig. 247 we have, 

ND = DE, BA = OA, KH = HG = As, etc. 

Therefore, 

As = FK—FH, 


And since by definition, As = s max . — s mln ., equation (237) may also 
be written: 

As = f(s„-s mIn .).(238) 

But s max . is the endurance limit corresponding to the range of 
stress As. If then s' r is used to denote the endurance limit, equa¬ 
tion (237) may be written 

s' r = s„—As.(239) 

9 As already noted, the more recent tests show that this value is nearly 
%s u , the higher ratio being due probably, in part at least, to more generous 
fillets and better surface finish. 
















318 


REPEATED LOADS. FATIGUE OF METALS 


It is frequently more useful to have the endurance limit 
expressed in terms of the ratio of the minimum unit-stress to the 



maximum unit-stress () than in terms of the range of stress. 


Thus, equation (237) may be transformed as follows: 


Smax. — As 




Therefore, 



2-q 1-W 


or since s max . is the endurance limit, s' r , for the given ratio, q, of 
minimum to maximum unit-stress we have 



(240) 


which will be called the Goodman-Johnson formula. 

The number of experiments made by Wohler to determine the 
effect of range of stress was not large; recent experiments, as 
already noted, indicate that the Goodman-Johnson formula gives 
results on the side of safety. 

Illinois Empirical Formula .—Since the endurance limits with 
completely reversed cycles of bending stress are now well estab¬ 
lished for a variety of steels, and can be found fairly easily from 
tests for any given steel (see Art. 156 for methods of testing) it is 
desirable to express the endurance limit of the steel for a given 
range of stress in terms of the endurance limit for completely 
reversed cycles of bending stress instead of the ultimate strength 
of the material. 

From a series of tests made in connection with the investiga¬ 
tion of the fatigue of metals at the University of Illinois, F. M. 
Howell found that the following empirical formula 10 expresses the 
relation between the endurance limit s' r of the material when sub¬ 
jected to any given maximum and minimum stresses in each cycle 

10 Bulletin 124, Engineering Experiment Station, University of Illinois. 





EFFECT OF RANGE OF STRESS 


319 


of bending stress and the endurance limit s f of a material when sub¬ 
jected to completely reversed bending stress: 



in which q is the algebraic ratio of the minimum stress to the 
maximum stress during a cycle of stress (for completely reversed 
stress q = 1.0). Equations (240) and (241), however, must be 
considered as tentative until more experimental data are available. 
(For a formula based on the energy theory see Bulletin 142, Engi¬ 
neering Experiment Station, Univ. of Illinois.) 

It will be assumed, tentatively, that with cycles of shearing 
stress the effect of range of stress is the same as with cycles of 
bending stress. This assumption seems decidedly on the side of 
safety since the results of the tests 11 now available indicate that, 
with repeated shearing stress, range of stress has considerably 
less effect on the shearing endurance limit than does range of 
stress on the endurance limit with repeated bending stress. 

Caution .—The limiting value of the unit-stress for any steel 
should never be taken greater than the yield-point even if the above 
formulas give a value of the endurance limit larger than the yield 
point. Thus, in the above equations the maximum value of s' r 
to be used is the yield-point of the material. That is, if s' r is 
found from equation (240) or (241) to be greater than the yield- 
point of the material the static strength of the member rather than 
the fatigue strength governs the design of the member. This 
fact should be clearly grasped since the value of s' r , in the above 
equations, for machine members that are subjected to cycles of 
stress in which the stress does not change sign is seldom less than 
the yield-point of the material. 

The Goodman-Johnson formula (or the Launhardt-Weyrauch 
formula which was similar to this) was formerly used extensively 
in designing bridge members. Static strength, however, will 
almost always govern the design of such members and the use of 
repeated stress formulas has practically been abandoned. It was 
frequently assumed that the use of a fatigue formula for bridge 
members was necessary to take account of the effect of impact 
loads. There was little justification, however, for using a fatigue 

11 Bulletin 142, Engineering Experiment Station, University of Illinois. 



320 


REPEATED LOADS. FATIGUE OF METALS 


formula for this purpose and now, since allowance is usually made 
for impact loads and certain secondary stresses, the effect of 
fatigue is properly neglected. 


ILLUSTRATIVE PROBLEM 


Problem 242.—The pulley, A, (Fig. 248) exerts a constant twisting moment 
of 4800 lb. in. on the shaft B. The total bending load P due to the tensions 

in the belt on the two sides of che pulley 
is 800 lb. The diameter of the shaft is 2 
in. and is mounted in flexible bearings. 
The shaft is made of cold-rolled steel for 
which the tensile ultimate strength is 
80,000 lb. per sq. in. and the tensile yield- 
point is 40,000 lb. per sq. in. The value 
of the endurance limit of the material is 
assumed to be unknown. Is the shaft well 
designed? 


B 


ir 


" <-12 > 
vp 




ii 

fZ 







Fig. 248. 


Solution.—Static Loading .—The stresses due to the loads, considering the 
loads to be static loads, are as follows: the shearing unit-stress due to the 
twisting moment is (see Art. 26), 


Tc 4800X1 


J 


^(2 y 

32 


= 3060 lb. per sq. in. 


The maximum fiber unit-stress at section CC due to the bending moment 
is (see Art.34), 

Me 9600X1 

s = ~r =—-—— = 12,240 lb. per sq. in. 

1 nz) 1 

64 

The maximum combined normal unit-stress is (see Art. 109), 
max. s' = *$+| Vs 2 +4s s 2 

= *12,240+*V12,240+4(3060) 2 
= 6120+6840 
= 12,960 lb. per sq. in. 
and the minimum normal unit-stress is 

min. s' = *s —*V / s 2 +4s s 2 


= 6120-6840 
= —720 lb. per sq. in. 





















ILLUSTRATIVE PROBLEM 


321 


As explained in Art. 109 max. s' and min. s' occur at the same point in the 
material, but on planes at right angles to each other. Further, when s is a 
tensile stress max. s' is a tensile stress and min. s' is a compressive stress, and 
when s is a compressive stress, max. s' is a compressive stress, and min. s r 
a tensile stress. 

The maximum combined shearing unit-stress is (see Art. 109). 
s'g = ±|Vs 2 +4s. 2 

= ±6840 lb. per sq. in. 

and this stress occurs at the same point where max. s' and min. s' occur but 
on each of two planes that bisect the angles between the planes on which max. s r 
and min. s' occur. 

Since a tensile and compressive working unit-stress of at least 16,000 lb. 
per sq. in. and a shearing working unit-stress of 10,000 lb. per sq. in could be 
used with static loading the shaft has ample resistance to static loads. 

Repeated Loads .—The number of repetitions of stress would be sufficient 
to cause rupture if the material is stressed above the endurance limit. The 
shaft has no abrupt change of section at the section, CC, of maximum stress 
but it will be assumed that the surface of the shaft is rather rough due to 
scratches caused by grit in the bearing, and it will be assumed that these 
scratches are in vertical planes and hence would cause localized bending stress 
only; the shearing stress, s s , being unaffected. 

Therefore, the value of s as found above must be increased. Thus, 
according to Table 4, 

s = 1.2X12,240 = 14,700 lb. per sq. in. 

if this value of s is used in the above equations the following values for max. s', 
min. s' and s' 8 are found 

max. s'= 7350+7950 

= 15,300 lb. per sq. in., 
min. s'= 7350—7950 

= —600 lb. per sq. in. 
s' 8 = ±7950 lb. per sq. in. 

Now the failure of the shaft due to repeated stress may be due to (a) the com¬ 
pletely reversed bending stress, (5) the completely reversed combined shearing 
stress, and (c) the combined normal stress that is not completely reversed. 

(a) The stress s (14,700 lb. per sq. in.) is a completely reversed bending 
stress. But the endurance limit of the material with completely reversed 
bending stress may be taken as 0.45 of the tensile ultimate strength, and hence: 
the above reversed stress is approximately 0.41 of the endurance limit, since 

W-.O.tt. 

0.45X80,000 




322 


REPEATED LOADS. FATIGUE OF METALS 


(6) The stress s' 8 (7950 lb. per sq. in.) is a completely reversed shearing 
stress. Now the shearing endurance limit of the material is (see Art. 127). 

(s r )s = s r X0.55 = 36,000X0.55 = 19,800 lb. per sq. in. 


Hence the reversed shearing stress is approximately 0.40 of the endurance 
limit since 


7950 

19,800 


0.40. 


(c) The stress s' varies in each cycle from min. s' ( — 600 lb. per sq. in., 
compressive stress) to max. s' (15,300 lb. per sq. in., tensile stress). Now 
for this range of stress the endurance limit of the material is 


1-k 


3 +q 


-s r . 


(see Art. 132) 


If the first expression is used we have: 


\8 0,000 


1 / — 600 N 
~2\15,300, 


40,000 
1 + 10.039 


40,000 ^ p er go, i n . 

1.020 


And hence the maximum stress in each cycle is approximately 0.39 of the cal¬ 
culated value of the endurance limit, since 


15,300 

39,200 


= 0.39. 


If the second expression is used we have: 


3+0.039 


0.45-80,000 


= 54,700 lb. per sq. in. 

Hence the maximum stress in each cycle is approximately 0.28 of the cal¬ 
culated value of the endurance limit. 

The critical stress, therefore, is the completely reversed bending stress, 
and its value is somewhat too high unless the uncertainty in the loads is rela¬ 
tively small. 

PROBLEMS 

243. A chrome-nickel heat-treated crank shaft of a single acting gas engine 
is subjected to repeated applications of a torsional shearing stress which varies 
approximately from zero to a maximum (q = 0); bending stresses may be 
neglected. If the shaft is 2 in. in diameter, what maximum torque should be 
allowed? Use values in Table 7. 

244. If in Problem 242 the distance from the action line of P to the sec¬ 
tion CC were 8 in. and the twisting moment were 6000 lb .-in., what stress 
would approach closest to the endurance limit? 











PROBLEMS 


323 


245. A special chrome-vanadium heat-treated steel having an ultimate 
tensile strength of 200,000 lb. per sq. in. and proportional limit of 150,000 
lb. per sq. in. is used in a helical spring on an automobile truck. The stress 
in the spring is mainly shearing stress (see Art. 90) and varies from a small 
value due to the weight of the truck to a maximum, (q = 0). The maximum 
stress is not constant, however, since the impact loads vary greatly. It may 
be assumed that the spring is subjected frequently to overload stresses. 
Estimate the endurance limit and find a working stress, assuming that the 
working stress should not be greater than one-fourth of the endurance limit. 

Ans . (s' r ) s = 74,400 lb. per sq. in.; s w — 18,600 lb. per sq. in. 


PART II. MECHANICAL PROPERTIES OF 
STRUCTURAL MATERIALS 


133. Introduction.—The effective use of materials in engineer¬ 
ing structures and machines for resisting loads requires a knowl¬ 
edge of (1) the loads to which the members of the structure or 
machine are to be subjected, (2) the relations between the loads 
and the stresses and deformations caused by the loads; these 
relations involve the dimensions and form of the body or member 
and are discussed in Part I, and (3) the mechanical or physical 
properties of structural materials. 

A knowledge of the properties of materials is needed for two 
purposes: ( a ) for use in establishing the relations under (2) and in 
the application of those relations to problems of design, and (6) 
for use in selecting materials best suited to the service require¬ 
ments. 

A discussion of the relations under (2) and a very brief discus¬ 
sion of the topics under (3) have been given in Part I. A more 
extended discussion under 3 (b), that is, of the properties needed 
in materials for load-resisting members, as used in various types of 
structures and machines, and of methods of measuring the prop¬ 
erties, is given in the following pages. 


324 



CHAPTER XV 


DEFINITIONS, METHODS OF MEASURING, AND 
SIGNIFICANCE OF MECHANICAL PROPERTIES 

134. Properties to be Considered.—A knowledge of the follow¬ 
ing properties is of special importance to the engineer in the use of 
materials in load-resisting structures and machines, and they can 
be measured quantitatively with a fair degree of satisfaction by 
means of mechanical tests. 

1. Strength 

(a) Static 

(b) Impact and Energy 

(c) Fatigue 

2. Ductility. 

On the other hand the properties, plasticity, malleability, and 
machineability are of importance mainly to the manufacturer in 
forming (rolling, forging, drawing, pressing, machining, etc.) 
the material for use in structures and machines, and their measure¬ 
ment is less definite than that of the properties mentioned above. 
However, these properties are closely connected with some of 
those mentioned above, as are also the general properties, flex¬ 
ibility and elasticity. 

135. Meaning of Strength.—By strength of a material is 
meant that property which enables the material to resist external 
forces or loads without incurring structural damage. By structural 
damage is meant stress and deformation in the material of a mem¬ 
ber that cause the member to cease to function properly in the 
structure or machine; it has different meanings under different 
conditions of use or service; for example, a stress or a deformation 
that would cause structural damage in the member of a building 
might not cause structural damage in a chain hoist or a coal-car 
frame, and, on the other hand, a deformation that would cause struc¬ 
tural damage in a lathe, planer, or other machine tool would not, 

325 


3. Stiffness 

4. Resilience 

5. Toughness 

6. Hardness 


326 DEFINITIONS, MEASURING, MECHANICAL PROPERTIES 


in general, cause damage in a a building, etc. Thus, the maximum 
usable strength of a material depends on the service in which the 
material is used, and accordingly it will be found desirable to 
have at least two measures l° f the maximum usable strength, 
namely, the ultimate strength and the elastic strength. However* 
no single quantity can adequately measure either the ultimate 
strength or the elastic strength of a material, for: 

(f) The strength of a material depends on the type of load¬ 
ing (static, impact or energy, and repeated) to be resisted. Thus, 
a cast-iron beam is stronger in resisting static loads than is a sim¬ 
ilar oak beam, whereas the oak beam is stronger than the cast-iron 
beam when the load applied is an impact or energy load (see Chap. 
XIII). Again, a bar of steel may resist a static load of a given 
magnitude without incurring measurable structural damage and 
yet it may rupture when subjected to a load of the same magnitude 
when applied a large number of times (see Chap. XIV). 

2. For any type of loading (static, impact and repeated) the 
strength of a material is different for different kinds of stresses 
(tensile, compressive and shearing) developed in the material. 
Thus the strength of cast iron, concrete, brick, and other brittle 
materials in tension and in shear are much less than their strengths 
in compression, and for some brittle materials including cast iron 
the shearing strength is greater than the tensile strength. Further* 
most ductile materials such as wrought iron, steel, soft brass, etc., 
are much weaker in shear than in tension and in compression. 

The strength of a material may, therefore, be considered in 
accordance with the following outline: 

Strength: 


1. Static; 


| Elastic 
1 Ultimate 


Tensile 

Compressive 

Shearing 


2. Impact and f Elastic 
Energy; j Ultimate 


Tensile 

Compressive 

Shearing 


f Tensile 

3. Fatigue; Ultimate <j Compressive 

[ Shearing 


Energy and fatigue strengths have already been discussed 
briefly in Chapters XIII and XIV, respectively. 


STATIC ULTIMATE STRENGTH 


327 


Static Strength 

136. Static Ultimate Strength.—The resistance that a material 
offers to a static (gradually applied) load, that is, the static strength 
of the material is measured in terms of the unit-stress (internal 
force per unit area) developed in the material. The static ultimate 
strength of a material is the unit-stress developed in the material 
by the maximum static load that the material can resist without 
rupturing. 

Method of Determining Ultimate Strength .—The static ultimate 
strength of a material is determined by testing a specimen of the 
material (or several specimens) in a testing machine (see Fig. 3 
for a common type of machine) which weighs the load applied to 
the specimen; the largest load applied before fracture occurs 
divided by the original area is taken as the measure of the ultimate 
strength. In determining the ultimate strength from a test, 
whether it be the tensile, compressive, or shearing ultimate 
strength, care must be exercised, particularly with brittle mate¬ 
rials, in securing an axial load (Art. 3), that is, a load which causes 
a uniformly distributed stress on the cross-sectional area of the 
test specimen, since the value of s, as found from the equation 

P 

s = —, 
a 

is the average unit-stress and not the maximum unit-stress devel¬ 
oped in the specimen if the maximum load P does not cause the 
stress to be distributed uniformly on the cross-sectional area. 

Experience has shown that the ordinary spherical-seated grips 
(see Fig. 249a) used on testing machines for tension and compres¬ 
sion tests do not insure that the load shall be axial; for ductile 1 
materials, however, the eccentricity introduced by such grips will 
not seriously affect the value of the ultimate strength. But, for 
brittle materials such as cast iron, cast aluminum alloys, etc., the 
eccentriticy may lower the value of the ultimate strength appre¬ 
ciably. In testing cast aluminum alloys it has been found 2 

1 The property of ductility is discussed in Art. 141, but for use in this 
article it will be sufficient to define ductility as that property which enables 
the materials to draw out or become plastic under load; brittleness is the lack 
of ductility. 

2 Report on the Materials of Construction used in Aircraft and Aircraft 
Engines, Aircraft Production Department of the Ministry of Munitions, 
British Government. 


328 DEFINITIONS, MEASURING, MECHANICAL PROPERTIES 


that unless special loading devices (see Fig. 249e) are used the 
ultimate strength obtained is frequently 30 per cent too low and 
that the low values are obtained consistently if a definite routine 
is followed in putting the specimens in the grips. 

Other tests 3 have shown that, even with approved methods of 
holding the specimen, the maximum unit-stress along an element 
or fiber of the specimen may be 10 per cent or more in excess of 
the mean unit-stress over the cross-sections. 

Form of Test Specimen .—The form of the test specimen also 
has an influence on the distribution of stress on the cross-section 
within the gage length. For relatively ductile materials the form 
of grips and of specimens shown in Fig. 249(c) and (d) give satis¬ 
factory results. In a long tension specimen the unevenness of 
stress caused by the gripping device has a better opportunity to 
become nearly uniform on the cross-section within the gage length 
than in a short specimen. ■ Therefore, spherical seated bearing 
blocks (Fig. 249a and b) are used, as a rule, only on short 
specimens. Short tension specimens are either threaded (Fig. 
249a) or have shoulders on the ends (Fig. 2496); the latter 
method is used particularly with specimens that are to be tested 
in a hardened condition after being heat treated. The loading 
device shown in Fig. 249(e) is used in testing very brittle material. 
(For compression and shearing test specimens see Art. 139 and 
140.) The significance of the ultimate strength of a material to 
the engineer is discussed in Art. 138. Values of ultimate strengths 
for various materials are given in the tables of Chapter XVI. 

137. Static Elastic Strength.—In many structures and 
machines structural damage to the material occurs at stresses 
much below the ultimate strength of the material, that is, the 
maximum usable strength of the material is not, as a rule, best 
measured by the ultimate strength. There is need, therefore, for 
a measure of the unit-stress at which structural damage begins 
or at which only a small amount of structural damage has occurred. 

The static elastic strength of a material is the maximum unit- 

3 “On the Stress Distribution During Tension Test,” Engineering (Lon.) 
Dec. 10, 1907, p. 796; Oct. 29, 1909, p. -593. For tension tests of specimens 
held in spherical-sea ted holders the ratio of maximum stress to mean stress 
averaged 1.165; for compressive tests of specimens with spherical seated 
bearing blocks the ratio averaged 1.059; for both series of tests the average 
was 1.101. 


STATIC ULTIMATE STRENGTH 


329 




_To Fit Gripping__ 

Device on Testing ^ 
Machine 


Test Specimen 


(e) 



Drive Fit 

Steel 

Ball 

Hardened 
Steel Plug 


/Machined at 
the Same 
Setting 




Fig. 249.—Methods of testing material in tension. 

























































































































































































































330 DEFINITIONS, MEASURING, MECHANICAL PROPERTIES 


stress that can be developed in the material without causing appre¬ 
ciable structural damage. The criterion for structural damage is 
somewhat indefinite, but, in general, structural damage, when 
caused by static loading , 4 is closely associated with plastic action 
in the material, and the unit-stress at which plastic action is con¬ 
sidered to begin or to have reached a measureable amount is 
usually taken as the measure of the static elastic strength of the 
material. In an actual test of a material (whether in tension, in 
compression or in shear) the determination of the unit-stress at 
which plastic action begins is subject to uncertainties, and hence 
a number of limiting stresses have been proposed and used as 
measures of the static elastic strength of a material; the various 
limiting stresses differing mainly by the amount of plastic action 
that is considered to be most significant or most convenient to 
measure. 

The following measures of the static elastic strength of a 
material are discussed below: proportional limit, yield point, 
elastic limit, Johnson’s apparent elastic limit, A.S.T.M. elastic 
limit, and proof stress. Further, although it is assumed, for con¬ 
venience, that the material is tested in tension, the discussion will 
also apply to a compression and a shearing test. Values of the 
static elastic strengths of various materials are given in the tables 
of Chapter XV. 

(a) Proportional Limit .—If a bar of mild steel is subjected to a 
gradually increasing axial load P (Fig. 250a), the unit-stress 

p e 

s, equal to —, and the unit-elongation (unit-strain) e, equal to 

will increase so that the unit-stress is proportional (within closely 
approximate limits) to the unit-strain (Fig. 2506), until a unit- 
stress called the proportional limit (sometimes called proportional 
elastic limit) is developed in the material: the proportional limit 
is represented by A'A in Fig. 250(6) and' (c). The curve of Fig. 
250(6) is called a stress-strain diagram and is obtained by plotting 
the unit-stresses and corresponding unit-strains as found from the 
test data. The unit-strains that occur before the proportional 
limit is reached, represented by the abscissa to the first part of 
the stress-strain curve in Fig. 250(6), are plotted to a larger scale 

4 Material may rupture when subjected to repeated loads without giving 
any evidence of plastic action, as is discussed in Chapter XIV and also in Art. 
155. 


STATIC ELASTIC STRENGTH 


331 


in Fig. 250(c) in order better to determine the modulus of elas¬ 
ticity (or the slope of the curve OA) and the proportional limit 
(or the point of tangency of OA and ABC), and also to bring out 
more clearly the form of the part ABC of the curve, since the strains 
corresponding to this part of the curve are closely connected with 
the beginning of structural damage. 

The proportional limit of a material, then, is the greatest unit- 
stress to which the material may be subjected without causing the 
unit-strain to increase at a faster rate than does the unit-stress. 



Fig. 250.—Stress-strain diagram of ductile steel; (6) the complete diagram, 
(c) the first part of diagram using a large scale for unit-elongations. 

The questions that now arise may be stated as follows: Is the 
proportional limit of a material, as determined from test data, 
the unit-stress at which plastic action (structural damage) begins ? 
Is it a reliable measure of the maximum usable stress for material 
as used in structures and machines? In order to answer these ques¬ 
tions the conditions or factors in the test which influence the value 
of the proportional limit must be investigated; and further, the 
extent to which these conditions influence the usefulness of the 
material in the structure or machine must be considered. 

Several of these influencing factors are discussed as follows: 

1. The proportional limit may be lowered appreciably by 
unevenness of stress in the fibers of the test specimen: 

(a) Unevenness of stress may be due to eccentricity of loading 
caused by the gripping devices, etc. For example, in Fig. 251 are 

















332 DEFINITIONS, MEASURING, MECHANICAL PROPERTIES 


shown the tension stress-strain curves obtained by plotting the 
strains measured on three independent gage lines (120° apart) 
against the corresponding average unit-stresses on the cross- 
section of the specimen, when the out-of-alignment of the jaws of 
the machine (the eccentricity of loading) was 0.035 in. These 
curves indicate the great unevenness of deformation (and hence of 
stress) due to eccentricity of loading. Such unevenness of stress 
tends to lower the proportional limit and to produce a long bend 
for the portion AB (Fig. 250c) in the stress-strain curve indicating 



Fig. 251.—Stress-strain diagrams showing unevenness of stress in specimen 
held in wedge grips when one pair of grips was 0.035 in. out-of-line. 


plastic action over a wide range of stress. The curve found by 
plotting the average of the strain readings on the three-gage lines 
is also shown in Fig. 251; this curve gives but little evidence of the 
uneven stretching of the specimen, and hence with the usual exten- 
someter, which averages the deformations on two or more gage 
lines, eccentricity of loading will not necessarily be detected by 
noticeable irregularities in the stress-strain diagram. 

(6) Unevenness of stress may be due to initial internal stresses 
caused by hardening steel, bronze, etc., by heat treatment or 
by cold working as in cold drawing and rolling. The stress-strain 
diagram for steel containing initial stresses is usually of the form 
shown in Fig. 252(a), and after the stresses are released by heating 
the steel mildly and cooling relatively slowly (tempering) the curve 
takes the form shown in Fig. 252(5); the increase in the propor¬ 
tional limit being large; in some cases much the same effect can 















































STATIC ELASTIC STRENGTH 


333 


be produced by subjecting the steel to several thousand repe¬ 
titions of low stress. 

2. The more sensitive the extensometer is, the lower on the 
curve will the proportional limit or point of tangency, A (Fig. 
250c), be detected, other conditions being the same. Thus, in 
one series of tests 5 the results given in Table 10 were obtained, 
three specimens being taken from each of three different bars of 
soft steel of supposedly the same quality. 

TABLE 10 


Smallest Unit-deformation 
Indicated by Extensometer 

Specimen 
from Bar 
No. 

Proportional 

Limit. 

Lb. per sq. in. 

Average Value for 
the Three Speci¬ 
mens in Each Group 

0.000,000,27.I 

1 

38,260 >| 


2 

32,670 

32,700 

l 

3 

27,160 J 


f 

1 

40,460 > 


0.000,000,5. 

2 

35,250 

34,480 

1 

3 

27,740 J 


r 

1 

41,320 > 


0.000,012,5. 

2 

34,590 

38,180 

1 

3 

38,620 J 



Thus, the average of the proportional limits of the specimens 
on which the most sensitive extensometer was used is 5480 lb. 
per sq. in. lower than that of the specimens on which the least 
sensitive (but commonly used) extensometer was employed. 

3. The value of the proportional limit of metals is also affected 
by the crystalline structure of the material. For example, metal 
having large or coarse grains due to slow cooling from a high 
temperature (as in the case of steel and brass castings, etc., or of 
rolled and forged steel on which the rolling or forging has been 
stopped at a temperature considerably above the critical tempera¬ 
ture) has a relatively low proportional limit since the stress-strain 
diagram is curved over a considerable distance along the curve as 

5 “ The Physical Significance of the Elastic Limit,” by H. F. Moore. Proc. 
Inter. Assoc, for Testing Materials, Vol. XXVIII. 











Unit - Stress in lb. persq, in. 


334 DEFINITIONS, MEASURING, MECHANICAL PROPERTIES 


indicated in Fig. 252(a), and any treatment which will refine the 
grain (such as heat treatment, or mechanical working of the metal 
while it is cooling to the critical temperature) may materially 
increase the proportional limit, as is shown in Fig. 253. 

4. The stress-strain diagrams for some brittle materials such 
as cast iron, concrete, etc., are curved practically from the start, 
similar to that shown in Fig. 252(a) (see Fig. 261), and hence the 
proportional limit for these materials is practically zero. 

5. When the stress-strain curve bends away from the straight 
line very gradually as in Fig. 252(a) and 253(a) the determination 



Fig. 252.—Effect of internal stress due to hardening by heat treatment on 
proportional limit of steel. 

of the point of tangency of the curve and the straight line (pro¬ 
portional limit) is subject to considerable variation and is also 
dependent upon the scales used in plotting the stresses and strains. 

Summary. —The value of the proportional limit of a material 
as found from a test is influenced (a) by local yielding of the speci¬ 
men (localized strains) due to a slight eccentricity of load, initial 
stresses, etc., particularly when determined with an extensometer 
that will detect very small deformations: ( b ) by the crystalline 
or internal structure of the material due to methods of cooling 
and of mechanical working, the value of the proportional limit 
for some coarse-grained brittle materials being practically zero. 

Is the Proportional Limit a Reliable Measure of the Maximum 
Usable Stressf The question now arises as to whether it is desirable 














































STATIC ELASTIC STRENGTH 


335 


to have a measure of the maximum usable stress that is influenced 
by the factors discussed above. It is undoubtedly true that the 
material in a member of a structure or machine is seldom if ever 
free from localized stresses and strains sufficient to cause local 
yieldings when the member is subjected to its working or design 
loads, these local yieldings, however, do not as a rule cause struc¬ 
tural damage to the member as a whole provided that the material 
has sufficient ductility to permit the local yielding to occur and 



Fig. 253. —Stress-strain diagrams of steel casting and rolled steel showing 
the effect of the hot-working, (a) Specimen from steel casting that 
cooled slowly after being poured. (6) Specimen of same chemical com¬ 
position as (a), steel rolled while cooling. 

thereby relieve the material of the high localized stresses. Hence, 
the maximum usable strength of a ductile material is likely to be 
greater than the proportional limit. Further, brittle material 
such as cast iron and concrete can be used safely when subjected 
to stresses above their proportional limits. 

It appears, therefore, that the proportional limit may be 
regarded as a reliable measure of the maximum usable strength 
of a ductile material provided that it is found from a specimen 
having a minimum of initial stress and that the specimen is tested 
with a minimum of eccentricity of loading so that the first yielding 


























336 DEFINITIONS, MEASURING, MECHANICAL PROPERTIES 


is also a rather general yielding of the whole specimen, as indicated 
by an abrupt type of stress-strain curve. 

Much of the hot-rolled and forged structural steel (and other 
hot worked ductile metals) which has been finished close to the 
critical temperature is fairly free from internal stresses and is fine 
grained; and, if tested with standard gripping devices (see Fig. 
249) and strain-measuring apparatus, the proportional limit is a 
fairly satisfactory measure of the beginning of real structural 
damage of such material. However, if the rolling has been 
stopped at too high a temperature as sometimes occurs with heavy 
rolled sections, plates, etc., the crystalline structure will be coarse 
and the stress-strain curve will be of the gradual type as shown in 
Fig. 252(a) and 253(a) instead of the abrupt type shown in Fig. 
252(5) and 253(6). Or, if the rolling has been finished at too low 
a temperature as is likely to occur with the thinner rolled-sections 
and with plates, initial stresses may be set up in the material, due 
to the cold working, which may cause the proportional limit to be 
less than the stress that would cause structural damage in the 
structure or machine in which the material is used. Further, 
for most hard steels and other brittle materials in general, the pro¬ 
portional limit is not a satisfactory measure of the static elastic 
strength. And, in any case, the uncertainties as to the real value 
of the proportional limit make desirable the measurement of the 
elastic strength by means of a unit-stress at which there occurs a 
small but measurable plastic deformation which is an indication of 
incipient general yielding of the specimen rather than by the pro¬ 
portional limit which is likely to be an indication of the beginning 
of local yielding that may not be indicative of structural damage 
under service conditions. Thus, the yield-point, Johnson’s 
Apparent Elastic Limit, etc., will now be considered as measures 
of the maximum usable strength of a material. 

(6) Yield-point .—The yield-point of a material is the unit- 
stress at which the material yields appreciably without an increase 
of load. In Fig. 250 the yield-point is represented by the ordinate 
to the horizontal line BC. Only ductile materials have yield- 
points. It is evident that structural damage to most structural 
and machine members has occurred when the primary stresses 
in the member reach the yield-point of the material, since the 
member will be permanently distorted. 

At stresses below the proportional limit the deformation is 


STATIC ELASTIC STRENGTH 


337 


practically all elastic, whereas at the yield-point the deformation 
is mainly plastic yielding. At stresses between the proportional 
limit and the yield-point (between points A and B, Fig. 250c) 
the deformation is partly elastic and partly plastic. If the change 
from elastic to plastic deformation is very gradual the ratio of the 
yield point to the proportional limit is large. This ratio is some¬ 
times used as a measure of the amount of internal stress in the 
material and hence as a measure of the degree of success in carding 
out various heat treatments of steel, since the proportional limit 
of a ductile steel free from internal stress and tested under an axial 
load is practically the same as the yield-point. 

Methods of Determining the Yield-jpoint. —The yield-point is 
usually determined in commercial testing by reading the load 
indicated by the poise on the beam of the testing machine (see 
Fig. 3) when the beam drops and remains down for some time as 
the specimen continues to stretch, thus indicating appreciable 
stretch without increase of load. It may also be found by use of 
dividers; the dividers are set to span a relatively short gage 
length on the specimen and are held on the specimen, and the 
least load that causes the gage length to become greater than the 
span of the dividers is noted and taken to be the yield-point load. 

The value of the yield-point is affected less than is the propor¬ 
tional limit by the conditions discussed under (a) above, but when 
determined by the “ drop of the beam ” the value may be raised 
from 3000 to 5000 lb. per sq. in., by the manipulation oi the oper¬ 
ator, by moving out the poise rapidly just as the yield-point is 
approached and thus taking advantage of the inertia of the beam. 
For a discussion of the determination of the so-called yield-point 
for non-ductile materials, see Art. 137(/). 

Is the Yield-point a Reliable Measure of the Maximum Usable 
Stressf The question now arises as to whether the yield-point, if 
carefully determined, is a reliable measure or criterion of the 
static elastic strength of a material to be used in structures and 
machines. The material in some structures and machines is not 
damaged if the stress developed is greater than the yield-point but 
for most structural uses the primary stresses in the members must 
be kept below the yield-point if the members are to perform their 
function in the structure. In such service and for ductile material 
the yield-point may be regarded, as the maximum usable strength 
of the material. 


338 DEFINITIONS, MEASURING, MECHANICAL PROPERTIES 


The yield-point, therefore, is a reliable measure of the minimum 
stress at which real structural damage becomes evident and hence 
of the maximum usable stress provided: (1) that the material is 
ductile enough to possess a real yield-point, (2) that it is care¬ 
fully determined, and (3) that the material is in such a condition 
(free from internal stress, etc.) that the change from elastic to 
plastic deformation is rather abrupt (Fig. 2536) instead of grad¬ 
ual (Fig. 253a). 

If a material is ductile, is free from internal stresses (such as 
most rolled and forged low carbon steel), and is loaded axially 
when tested, the proportional limit and the yield-point are prac¬ 
tically the same value, and either one is a satisfactory measure of 
the static elastic strength; whereas, for metals that have been so 
treated as to produce a coarse-grained structure or to produce 
internal stresses (as in hardening by heat treatment or in cold 
working) there may be a wide difference between the values of the 
proportional limit and yield-point, and under such conditions 
neither the proportional limit nor the yield-point is as reliable a 
measure of the maximum usable strength of the material as is 
frequently desired. Further, neither of them is a satisfactory 
measure of the elastic strength of most brittle materials. Hence f 
there is need of a method of determining a stress between the pro¬ 
portional limit and the yield-point at which the deformation is 
partly elastic and partly plastic, with sufficient plastic action to 
be readily measurable but sufficient only to indicate incipient 
general yielding of the material. Several such measures of the 
elastic strength are discussed under (d), (e), and (/) below; before 
discussing these, however, the elastic limit of a material will be 
considered. 

(c) Elastic Limit .—Permanent deformation or set is funda¬ 
mentally the best evidence of plastic action or of the breakdown of 
elastic action in the material. The maximum unit-stress that can 
be developed in the material without causing a permanent set is 
called the elastic limit of the material. It may be found by apply¬ 
ing a small load to the specimen and then releasing the load and 
noting whetherThe extensometer reading returns to zero; if there 
is no permanent set a larger load can be applied and so on, until a 
slight permanent set is indicated by the extensometer. If still larger 
loads are applied and then released, the sets will be found to increase 
similar to that indicated by the stress-set curve in Fig. 254(a). 


STATIC ELASTIC STRENGTH 


339 


The value of the elastic limit as found from tests is affected 
by the same conditions (eccentricity of load, sensitiveness of exten- 
someter, initial stresses, etc.) as is the value of the proportional 
limit, and permanent sets may frequently be detected at low 
stresses particularly with sensitive extensometers but such sets, 
due largely to initial and localized stresses, although present in 
the members of many structures and machines, do not seriously 
affect the strength of the member as a whole, provided that the 
material is relatively ductile. And hence such preliminary sets 



Eig. 254.—Methods of determining Fig. 255. — Tensile stress-strain 
the maximum usable strength of a diagram for rubber; maximum 
ductile material. unit-stress in cycle about 0.70 of 

ultimate strength. 

are, as a rule, not a reliable measure of structural damage when the 
member is subjected to static loads. 

For ductile metals the value of the proportional limit and 
elastic limit as found from tests are practically equal. There is, 
however, no fundamental relation between the proportional limit 
and the elastic limit of a material. In other words, there is noth¬ 
ing in the property of elasticity which requires that unit-stress 
shall be proportional to unit-strain; a material that has a curved 
stress-strain diagram might retrace that curve on release of the 
stress and hence have perfect elasticity. In fact, the stress-strain 
curve for rubber is a reversed curve (Fig. 255) and on release of the 
load (if the load is not too large) the material shows no permanent 
deformation although the curve obtained during release of load 
is not coincident with the curve obtained during application of 
load, that is, there is a hysteresis loop. 


























340 DEFINITIONS, MEASURING, MECHANICAL PROPERTIES 


However, although there is no fundamental relation between 
the proportional limit and the elastic limit, tests indicate that for 
all wrought structural metals and for practically all structural 
materials a deviation from a straight stress-strain diagram is 
accompanied by plastic deformation or set. For most structural 
materials, therefore, the proportional limit serves the same pur¬ 
pose as does the elastic limit and is much easier to determine. 

(i d) Johnson’s Apparent Elastic Limit. —In order to avoid the 
objections to both the proportional limit and the yield-point, as 
discussed under ( a ) and (6) above, J. B. Johnson 6 used a unit- 
stress called the apparent elastic limit which is the unit-stress at 
which the deformation increases, with respect to stress, at a rate 
50 per cent greater than that at zero (or small) stress. It is found 
by drawing a line OD (Fig. 254) on the stress-strain diagram 
which has a slope 50 per cent greater than that of the straight 
part of the curve below the proportional limit, and then drawing 
a line D'C' parallel to OD and tangent to the stress-strain curve; 
the ordinate to the point of tangency is the unit-stress at which 
the deformation is increasing, with respect to stress, 50 per cent 
greater than at any stress below the proportional limit. 

The value of 50 per cent increase in slope is arbitrarily selected, 
the purpose being merely to select, by a convenient method, a unit- 
stress at which a small but measurable deviation from the straight 
stress-strain line which is indicative of the beginning of a general 
plastic yielding of the specimen. 

A unit-stress selected by the same method but by using a value 
of 100 per cent increase in slope, instead of 50 per cent increase 
has been called the useful limit point. 

The method of determining Johnson’s apparent elastic limit is 
applicable to nearly all forms of stress-strain curves obtained from 
structural materials and the value obtained is probably the best 
single measure of the maximum usable stress (static elastic strength) 
of a structural material. 

(e) A.S.T.M. Elastic Limit. —As stated under (6) above, the 
yield-point is usually a satisfactory measure of the elastic strength 
of a ductile material, provided that reasonable care is taken in 
securing an axial load in testing, and that the material is reasonably 
free from internal stresses, etc., so that the yield-point is sharply 
defined. Many materials, such as heat-treated or cold-rolled 
6 Johnson’s Materials of Construction, 5th Edition. 


STATIC ELASTIC STRENGTH 


341 


medium-carbon and alloy steels used for axles, etc., however? 
do not have a sharply defined yield-point, and for such materials 
a measure of static elastic strength is found, particularly in com¬ 
mercial tests which determine acceptance or rejection of the 
material, by the method prescribed in some of the specifications 
of the American Society for Testing Materials, described as 
follows: “ The elastic limit called for by these specifications (for 
heat treated axles and shafts, cold-rolled axles, etc.) shall be deter¬ 
mined by an extensometer reading to 0.0002 in. The exten- 
someter shall be attached to the specimen at the gage marks and 
not to the shoulders of the specimen nor at any part of the testing 
machine. When the specimen is in place and the extensometer 
attached the testing machine shall be operated so as to increase 
the load on the specimen at a uniform rate. The observer shall 
watch the elongation of the specimen as shown by the extensometer 
and shall note, for this determination, the load at which the rate 
of elongation shows a sudden increase.” 

The stress found by this method, as a rule, will be represented 
by a point on the stress-strain curve somewhat above Johnson’s 
apparent elastic limit and hence should be considered perhaps 
as an accurate and conveniently made determination of the yield- 
point; it has decided advantages over the method of determining 
the yield-point by the “ drop of the beam ” or by dividers par¬ 
ticularly for the class of material that exhibits a gradual increase 
in plastic yielding. However, for high-carbon steel, cast iron, and 
other metals of a like degree of brittleness, the “ sudden increase ” 
referred to above is not well marked, and for such material this 
determination tends to yield unsatisfactory results. 

(/) Proof Stress .—As has already been stated the yield-point 
is defined to be the unit-stress at which deformation occurs without 
increase of load and that one of the methods of determining the 
yield point is to obtain the load (and hence the unit-stress) at 
which a distinctly visible increase occurs in the distance between 
gage points on the test specimen as observed by using dividers. 
Now such a visible increase in deformation may be observed in 
material which, according to the definition has no yield-point, 
that is, the deformation observed is elastic deformation; this is 
particularly true in testing high-strength steel, when using a rel¬ 
atively long gage length (4 in. or over). Such a use of the yield- 
point is considered objectionable since the maximum usable stress 


342 DEFINITIONS, MEASURING, MECHANICAL PROPERTIES 


may be greater than the value found in this way. Hence a stress 
called the proof stress has been proposed and adopted, in British 

specifications, for certain classes of ma¬ 
terial as a substitute for the yield-point. 
The proof stress is a unit-stress applied 
for a short time (fifteen seconds, say) 
which must not produce a permanent 
set of more than a specified amount (§ 
per cent, say) when the stress is re¬ 
moved; that is, a permanent set less 
than the specified amount is not con¬ 
sidered to be evidence of structural 
damage. Thus, in Fig. 256 the proof 
stress for the material is represented by 
the ordinate to the point P on the stress- 
strain diagram. 

QUESTIONS 

1. Does any single quantity measure adequately either the elastic strength 
or the ultimate strength of a material? If not, why? 

2 . What is meant by “ structural damage ”? 

3 . What do static elastic strength and maximum usable stress mean? 

4 . Name several measures of the static elastic strength of a material. 

5 . State and discuss several conditions which affect the value of the 
proportional limit as found from a test. 

6. Should a measure of static elastic strength be influenced by the above 
conditions? 

7 . State several causes of unevenness of stress in a tension test specimen. 

8. How does unevenness of stress affect the form of the stress-strain dia¬ 
gram? 

9. Under what conditions is the proportional limit a reliable measure of 
the static elastic strength of a material? 

10 . Define yield-point of a material. 

11 . Of what use is the ratio of the proportional limit to the yield-point? 

12 . What advantages has the yield-point as a measure of static elastic 
strength over the proportional limit? Under what conditions are the values of 
the proportional limit and yield-point approximately equal? 







SIGNIFICANCE OF ULTIMATE AND ELASTIC STRENGTHS 343 


13. What kinds of material have no yield-point? 

14. Define elastic limit. Is there any fundamental relation between 
the elastic limit and proportional limit of a material? Is there any relation 
as found from experiment? If so, what relation? 

15. Define Johnson’s apparent elastic limit. 

16. What are the advantages of Johnson’s elastic limit as a measure of 
elastic static strength of a material? 

17. Define the A.S.T.M. elastic limit. For what class of materials is it 
specified? Why? 

18. What is the proof stress? Why has it been used in preference to the 
yield-point? 

138. The Significance of the Ultimate and Elastic Strengths.— 

The material in some structural and machine members are fre¬ 
quently subjected to stresses greater than those proposed above as 
measures of the elastic strength of the material as, for example, 
freight-car frames, hoisting cables, ball bearings, etc.; the plastic 
flow of the material which accompanies these stresses do not 
necessarily destroy the usefulness of these members although 
the “ life ” of the members may be reduced because of such use. On 
the other hand, the distortion of the members in many (most) 
structures when stressed, as a whole, above the elastic strength 
of the material will destroy the usefulness of the structure or 
machine; this is true of permanent buildings, bridges, railroad 
rails, machine frames, etc. 

If, then, members in their normal service are to be subjected to 
stresses above the elastic strength of the material the ultimate 
strength may be the maximum usable strength of the material. 
Further, in the case of brittle materials, which have no well-defined 
elastic strength, the ultimate strength is usually a fairly satis¬ 
factory measure of the maximum usable strength of the material 
regardless of the use of the material in the structure or machine, 
since for such material the deformation accompanying any stress 
below the ultimate strength is mainly elastic and hence the ulti¬ 
mate strength is also the elastic strength. 

However, for members in most structures and machines the 
elastic strength of the material is the maximum usable stress for 
the material. And, in all cases, the working stress is considerable 
less than the maximum usable strength, the value of the working 


344 DEFINITIONS, MEASURING, MECHANICAL PROPERTIES 

stress depending on the conditions and uncertainties discussed 

in Art. 7. 

But even though the elastic strength of a material is the max¬ 
imum usable strength, there are two very important reasons why 
the value of the ultimate strength of material used in force-resist¬ 
ing members should be known: 

(1) In order to give a measure of the reserve strength of the 
member, which it brought into play in resisting abnormally large 
overloads due to accident and unforeseen conditions, as for exam¬ 
ple, a derailment on a bridge, an excessive settlement of a founda¬ 
tion of a building or bridge, a tornado, the running of a testing 
machine after the specimen has been removed and until the 
moving head comes in contact with the weighing table, etc. In 
such cases, although the structure or machine may have its use¬ 
fulness temporarily destroyed, nevertheless, if the material has a 
large reserve strength (a large insurance factor against total col¬ 
lapse) much less damage may be done, particularly where loss of 
life is involved, than if the reserve strength is lacking and hence 
allowing collapse or rupture to occur. (This topic is discussed 
also in Art. 143 and 148.) 

(2) But, even if the loads on a member do not cause stresses, in 
the member as a whole, that are greater than the elastic strength 
of the material, there may be localized stresses at various points 
in the member that are considerably above the yield-point of the 
material; high localized stresses frequently occur at sudden changes 
of sections such as at the roots of threads, at the edges of the 
holes in the plates of a riveted joint, etc.; they are also caused 
by straining the members during erection in order to make them 
line up, or by heat treatment or cold working, etc. And, although 
only a small part of the total volume of the member may be over¬ 
stressed, still if the material of this small volume does not have a 
considerable reserve strength it will rupture and may be the initial 
cause of the rupture of the whole member. In fact stresses (both 
local and general) greater than the elastic strength of the mate¬ 
rial are by no means uncommon in many types of structures that 
are supposedly subjected to relatively low working stresses. The 
spalling off of paint frequently noticed on steel members in build¬ 
ings and other structures indicates the great need of “ reserve 
strength,” and at the same time suggests that in some classes of 
work the reserve strength of the material is too frequently and too 


STATIC COMPRESSIVE STRENGTH 


345 


confidently relied on to resist secondary stresses not taken into 
account in proportioning and erecting the structure or machine. 

Summary: The ultimate strength is of value in selecting suit¬ 
able working stresses for brittle material such as cast iron, concrete, 
rope, canvas, hard drawn copper, etc., since for such materials the 
ultimate strength, if determined carefully (see Art. 136), is prob¬ 
ably the best measure of the maximum usable stress. In struc¬ 
tures made of ductile material, however, the elastic strength must, 
as a rule, be considered to be the maximum usable strength of the 
material. But, even though the working stress as well as the 
selection of the material is based largely on the elastic strength, 
the ultimate strength, particularly in its relation to the elastic 
strength, is of great importance as a measure of the reserve strength 
of the material; that is, the strength which becomes available in 
resisting unforeseen overloads, excessive localized and secondary 
stresses, etc., thereby adding an insurance factor against total 
rupture or collapse of the structure or machine. The ratio of the 
elastic strength to the ultimate strength is frequently called the 
elastic ratio; in specifications for structural steel this ratio usually 
is required to be at least one-half, the elastic strength being meas¬ 
ured by the yield-point. 

Further, test results show that a value of approximately one- 
half of the ultimate static strength is a fairly satisfactory measure 
of the fatigue strength of wrought ferrous materials when the 
repeated stress is completely reversed, that is, when the stress 
varies from a tensile to an equal compressive stress (see Art. 155 for 
further discussion). 

QUESTIONS 

19. For what kind of a material and under what conditions may the ulti¬ 
mate strength be regarded as the maximum usable strength of a material? 

20. What is meant by reserve strength of a material? Why is reserve 
strength of importance even when the material is used in a structure that 
would be damaged by stressing the material above the elastic strength. 
Define elastic ratio. 

139. Static Compressive Strength.—Although the preceding 
discussion has referred mainly to the tensile strength of materials, 
the statements also apply with few exceptions to the compressive 
strength of the materials. There are, however, additional facts 
that should be considered in connection with compressive strength. 


346 DEFINITIONS, MEASURING, MECHANICAL PROPERTIES 


For most structural compression members (columns) made of 
ductile material the yield-point must be regarded as the ultimate 
strength of the material since the plastic yielding accompanying 
the yield-point allows the member to bend, after which rupture 
or collapse may soon follow (see Art. 92). Now, if a structure 
which contains both tension and compression members such as a 
building, bridge, crane, etc., and which is made of ductile material 
(structural steel, etc.), for which the ultimate strength in tension 
and compression are approximately the same, is to be designed, 
and if the same working stress is used for compression members as 
is used for tension members, the resistance of the tension members 
to total collapse will be greater than that of the compression mem¬ 
bers. In other words, the reserve strengths of the tension 7 
members will be greater than that of the compression members, 
and since the reserve strength of the whole structure is likely to 
be that of the compression members it is evident that, if the reserve 
strength of the structure is of real importance (as it is in most struc¬ 
tures), the working stresses for the compression members should 
be lower than for the tension members. This principle is recog¬ 
nized in city building ordinances and other building specifications 
by specifying lower 8 working stresses for compression members 
than for tension and flexural members. 

Further, the small local yieldings, which lower the proportional 
limit (see Art. 137a) but which in a tension member may not be 
indicative of structural damage may be the beginning of real 
structural damage in a column, since local yielding is likely to be 
followed by bending of the whole column, and hence for columns 
the measure of the elastic strength of the material needs more 
careful consideration than for members used in tension or flexure; 
the conditions (amount of rolling of metals, of moisture in timber, 
of internal stresses in metals, etc.), which affect the elastic strength 
should be considered in determining working stresses. 

Form of Test Specimen. —In determining the compressive 

7 The reserve strength of large built-up tension members is not as large in 
comparison with compression members as might be expected from the results 
of small test specimens, due to the unevenness of stress in the component parts 
of the built-up member. Further, the tensile strength of full-sized solid mem¬ 
bers may be only about 0.85 of the tensile strength of test specimens. (See 
Trans Am. Soc. Civ. Eng., Vol. 61, p. 191; also Eng. News, July 5, 1906.) 

8 See footnote 17 of Chap. XI. 


STATIC SHEARING STRENGTH 


347 


strength of a material the specimen should not be long enough to 
allow it to bend, and yet it should be long enough so that the 
unevenness of the pressure on the ends of the specimens will not 
cause noticeable unevenness of stress on cross-sections in the 
central portions of the specimen, that is, in the gage length. A 
cylindrical specimen having a length two to four times its diam¬ 
eter is satisfactory, provided the ends are machined smooth in the 
case of metals or imbedded in plaster of paris if the specimen is 
concrete, stone, etc. Thus, a cast-iron or steel cylinder 1 in. by 
3 in., and a concrete cylinder 6 in. by 12 in. are in common use. 

140. Static Shearing Strength.—The shearing ultimate strength 
of a material may be found by testing the material in direct shear 
in various ways as indicated in Fig. 257. The shearing elastic 
strength (proportional limit, Johnson’s apparent elastic limit, 
yield-point, etc.) is not found from a direct shearing test because 
of the difficulty of measuring the deformation. The shearing 
elastic strength is found from a torsion test of a cylindrical speci¬ 
men (for one type of a torsion testing machine see Fig. 258). The 
stresses on each cross-section of the specimen are shearing stresses 
and the maximum stress may be calculated from the twisting 
moment which is applied and measured by the machine (Fig. 258), 
and the shearing unit-strain may be calculated from the value of 
the angle of twist; the angle of twist, 0, for a gage length, l, being 
measured by the scale and pointer attached to the specimen as 
indicated in Fig. 258. Therefore, a shearing stress-strain diagram 
may be plotted. 

Thus if a gradually increasing twisting moment, T, is applied 
to a solid cylindrical bar of ductile material, such as structural 
steel, the stress-strain curve will be as shown in Fig. 259; the 
shearing unit-stress, s s , and the shearing unit-strain, e s , as shown 
in Art. 26 and 28, are found from the equations 

Tc , cd 
S s = -j and e s = y. 

From this curve the approximate values 9 of the shearing pro¬ 
portional limit (P, Fig. 259), Johnson’s apparent elastic limit 
(E, Fig. 259), yield-point (F, Fig. 259), etc., may be found by the 
same methods as were discussed in Art. 137 in connection with a 
tension test. However, in a torsion test it is usually more con- 

9 See below for reasons why these values are approximate. 


348 DEFINITIONS, MEASURING, MECHANICAL PROPERTIES 


venient to draw a torque-angle of twist (T—d) curve; this curve 
will have the same form as that given in Fig. 11, since c, J, and l 
depend only on the dimensions of the specimen which are con- 



(a) 


Test for Rivets and Bars 




Fig. 257.—Methods of testing material in direct shear. 

stant. The torque, T, at the proportional limit or yield-point, 
etc., may then be found and the corresponding unit-stress calcu¬ 
lated from this torque. 






























































































































































STATIC SHEARING STRENGTH 


349 



Fig. 258.—Torsion testing machine. By turning the handle the large gear 
is rotated, thereby exerting a twisting moment on the right-hand end of 
the specimen. This moment twists the specimen and deflects the pen¬ 
dulum to which the left-hand end of the specimen is attached; the 
deflection or swing of the pendulum is made to measure the value of 
the twisting moment. 














350 DEFINITIONS, MEASURING, MECHANICAL PROPERTIES 


True Measure of Shearing Elastic Strength. —Since the shearing 
unit-stress in a bar subjected to torsion, increases directly 
with the distance from the center of the bar, the fibers at the 
surface reach their proportional limits first, but the increase in the 
angle of twist due to the yielding of these surface fibers will be too 
small to be measured by the strain-measuring apparatus. Hence, 
the proportional limit, as found from the stress-strain (or torque- 
angle of twist) curve for a solid bar will not represent the real 
proportional limit of the material. By using hollow thin-walled 
cylindrical torsion specimens the proportional limit indicated by 



Fig. 259. —Shearing stress-strain diagram for ductile steeJ. 

the curve will be very close to the true value for the material 
since nearly all the material reaches the proportional limit at the 
same time. 

Tests 10 have shown that the true shearing elastic strength of 
ductile and semi-ductile steels is about eighty-five hundredths 
(0.85) of the elastic strength found from a test of solid cylindrical 
torsion specimen. 

Relation between Shearing and Tensile Elastic Strengths .— 
Tests 10 have shown also that the shearing elastic strength of duc¬ 
tile and semi-ductile steels does not vary much from six-tenths 
(0.6) of the tensile elastic strength, whether the elastic strengths 
are measured by the proportional limits, the useful limit points, or 
the yield-points. 

10 Bulletin No. 115, Engineering Experiment Station, University of Illinois. 





DUCTILITY 


351 


QUESTIONS 

21. State reasons why the allowable working stresses in columns made of 
ductile material should, in general, be lower than in tension and flexural 
members. 

22. Why is the shearing elastic strength of a material found by means of a 
torsion test instead of a direct shearing test? 

23. Why is a hollow cylindrical torsion specimen required, in order to 
obtain an accurate value of the shearing proportional limit of a material? 


Ductility 

141. Definition of Ductility.—Ductility is that property of a 
material which enables it to acquire large permanent deformation 
and at the same time develop relatively large stress. It is closely 
associated with the properties plasticity and malleability since plas¬ 
tic and malleable materials can be worked (hammered, pressed, 
etc.) into various forms and will retain the forms impressed upon 
them; they acquire, therefore, large permanent deformations but 
the resistance that the material offers to the external forces im¬ 
pressed on it while being deformed is either relatively small or of 
secondary importance. Thus, lead and gold can be hammered 
into thin sheets and are said to be malleable; putty can be formed 
into various shapes with very little pressure and is said to be 
plastic; low carbon steel acquires a relatively large permanent 
elongation when subjected to a tensile load and maintains its 
resistance to the load and is, therefore, said to be ductile. There 
is, however, no sharp line of demarkation between the properties, 
ductility, plasticity, and malleability, but it is important to note 
that ductility is a property of a load-resisting material. 

142. Measure of Ductility.—There is no absolute quantitative 
measure of ductility. For comparative purposes the most fre¬ 
quently used measures are the values of the percentage of elonga¬ 
tion and the percentage of reduction of area of a specimen tested 
in tension. The percentage of elongation is found by dividing the 
increase in the gage length after rupture has occurred by the 
original gage length and multiplying the result by 100. The per¬ 
centage of reduction of area is found by dividing the difference 
between the area of the original and ruptured cross-sections by the 
original cross-section and multiplying the result by 100. 


352 DEFINITIONS, MEASURING, MECHANICAL PROPERTIES 


After the maximum load on the specimen of a ductile material 
has been reached the specimen begins to “ neck down ” and the 
load on the specimen decreases as the necking down continues 
(see Fig. 250). As indicated in Fig. 260, the deformation per unit 
length is very much greater at the neck-down portion than else¬ 
where. Therefore, the percentage elongation means little as a 
measure of ductility unless the gage length over which the defor¬ 
mation is measured is also specified. Thus a material that has a 
percentage of elongation of 32.5 in 8 in. might have a percentage 
of elongation of 60 in 2 in. (see Fig. 260). 


--- 

^_ n" _^ 


r-" ' 




L* _ q " _ 





T 

1 — 4 c 

i'l 











Specimen before pulling 


Specimen after pulling " v, ~ £ 











<2", 100# Elong> 

< -4.8" 60# Elong;- >■ 

7_s" 46F.Inner 


-> 


Q K 7 Flnncr ^ 

^ 10.6,' 32.5# Elong. 


Fig. 260.—The influence of the gage length on the percentage of elongation. 


Since during the “ necking down ” the load falls off rapidly 
and hence the material is in the process of rupturing, and since 
“ necking down ” should never occur in a member in a structure, 
there is justification for omitting the deformation that occurs dur¬ 
ing “ necking down ” in obtaining the percentage of elongation 
as a measure of ductility. Thus, according to this view, the duc¬ 
tility would be measured by the length OE (in Fig. 250). In 
commercial testing, however, the added time and expense involved 
in deducting the elongation due to the necking down has pre¬ 
vented the adoption of this method for measuring ductility. The 
percentage elongation, therefore, as usually found, includes the 
deformation due to the necking down (EF, Fig. 250) and also the 
small amount of elastic deformation (OA ', Fig. 250). 









































SIGNIFICANCE AND NEED OF DUCTILITY 


353 


Need for Other Methods .—Various grades of steel and of copper 
alloys differ widely in the way they stretch, some contracting to a 
narrow neck and some stretching uniformly and breaking with 
very little necking down; for such material the percentage of reduc¬ 
tion of area in addition to the percentage of elongation is some¬ 
times desired. Further, specimens of thin sheets of metal fail 
with almost no elongation, due to tearing from one side, and the 
ruptured area of such specimens is difficult to measure, and hence 
a bend test is frequently used as an indication (rough measure) 
of their ductility. 

This test consists in bending the specimen through 180° 
without causing cracking of the material on the outside of the bent 
portion, the radius of the bend being zero for thin (less than f in.) 
specimens of ductile material, and about equal to the thickness 
of the specimen for thicker specimens. 

143 . Significance and Need of Ductility.—Values that are 
considered to be satisfactory for the percentage of elongation and 
percentage of reduction of area for various materials are deter¬ 
mined mainly by the values obtained from material that is known 
to be approximately the best that can be produced, by the methods 
in use. In other words, it is practically impossible to state that a 
material must have a given percentage of elongation in order to be 
satisfactory for a given use. A value of 15 per cent elongation 
might be considered very satisfactory for a high carbon steel, 
whereas 25 per cent might be considered to be unsatisfactory for a 
low carbon steel, etc. Thus the values of the percentages of elonga¬ 
tion and reduction of area are a help in indicating to the engineer 
whether the material has been manufactured properly, in addition 
to being a measure of a property of the material. 

Why is Ductility Needed in Load-resisting Materialf —Is duc¬ 
tility of importance in material of members of structures and 
machines whose function or usefulness would be destroyed if the 
deformation of the member as a whole were large enough to make 
use of the ductility of the material? 

The answer to this question is closely connected with the dis¬ 
cussion of “ reserve strength ” in Art. 138 and also with the dis¬ 
cussion of “ toughness ” in Art. 147 and 148. Ductility is needed 
in the materials of structures and machines for the following 
reasons: 

1. To help prevent the destruction or collapse of the structure 


354 DEFINITIONS, MEASURING, MECHANICAL PROPERTIES 


due to excessive overloads, particularly impact loads, caused by 
accident, etc. This will be discussed more in detail under “ tough¬ 
ness/’ Art. 147. 

2. To relieve localized stresses in members by allowing the 
material to yield locally; this redistributes the stress, without 
causing any appreciable deformation of the member as a whole; 
thus the yielding prevents local failure of the material which in 
many cases would be the initial failure leading to the collapse of 
the whole member. These local stresses may be due to: 

(а) Loading that is different from that assumed in the design, 
and to secondary stresses not considered in the design; for example, 
the distribution of the load to the rivets in a riveted joint in design 
is usually assumed to be uniform, whereas the loads on the rivets 
are known to vary widely; the settlements of foundations fre¬ 
quently cause large changes in the loads on the members in the 
superstructure; the localized or secondary stresses (not considered 
in the design) in the eye of a steel eye-bar due to the bending action 
in the eye is sometimes sufficient to cause flaking off of the paint, 
etc. 

(б) Abrupt changes of section, such as corners in crank¬ 
shafts, roots of threads, key-ways, etc., at which high stresses 
always exist. 

(c) Non-homogeneity of the material due to a segregation 
in the ingot, and non-metallic inclusions and other defects in 
the case of steel; knots in timber; etc. 

(d) Straining members of structures during fabricating and 
erecting. 

Ductility, therefore, is of great importance in materials of 
most structures and machines and the amount of ductility is 
usually made as great as possible consistent with adequate strength; 
for, in general, the ductility of steel and other metals is low if the 
strength is high. 

Values of percentage of elongation for various materials are 
given in the tables of Chapter XVI. 

QUESTIONS 

24. Define ductility. What quantities are used as a measure of ductility? 

25. Why should the gage length be specified in giving the percentage of 
elongation? 


STIFFNESS 


355 


26. Why is ductility desired in a force-resisting member when, in order to 
make use of this property in the member as a whole, the member must take 
large permanent deformations and hence fail to fulfill its function in the struc¬ 
ture? 

27. State several causes of localized yielding in members of structures, and 
give illustrations. 

28. Is there any way of deciding how much ductility is required in material 
for a given use? 


Stiffness 

144. Definition and Measure. —The property of stiffness of a 
material is the rate at which the stress in the material increases 
with the strain. Hence, the stiffness of a material, corresponding 
to any unit-stress, may be measured by the slope of a line drawn 
tangent to the stress-strain curve at the point representing that 
unit-stress. 

The stiffness of steel, or any other material that has a straight 
stress-strain diagram up to the proportional limit, is constant at all 
stresses below the proportional limit and is measured by the mod¬ 
ulus of elasticity of the material ^2? = ^, Art. 5^, that is, by the rate 

of increase in unit-stress with unit-strain. Stiffness, therefore, is 
measured in the same units as is unit-stress (pounds per square 
inch) since e is the ratio of a length to a length and hence is 
merely a number. 

As shown in the tensile stress-strain diagram of mild steel in 
Fig. 250, the stiffness of mild steel when subjected to a tensile 
stress represented by H (about half-way between the yield-point 
and the ultimate strength) is very much less than at stresses below 
the proportional limit. At the yield-point the stiffness is zero. 
Further, it will be noted that Johnson’s apparent elastic limit 
(Art. 137d) is the unit-stress at which the stiffness of the material 
is 50 per cent less than at stresses below the proportional limit. 

Since in most structures the primary stresses developed are less 
than the elastic strength of the material the modulus of elasticity 
measures the stiffness of the material as used in such service, and 
hence is the most important measure of stiffness for design purposes. 
However, the low value of the stiffness of the material at points of 
high localized stress may be of importance, as for example, in the 
resistance of columns in which the stiffness of the material is an 


356 DEFINITIONS, MEASURING, MECHANICAL PROPERTIES 


important factor and in which localized stresses have a relatively 
large influence. 

For material that has a curved stress-strain diagram (Figs. 
261 and 262a), the modulus of elasticity is frequently taken as the 
slope of the tangent to the curve at zero stress, and is called the 
tangent or initial modulus. Thus, in Fig. 262(a) the slope of the 
line OA is the tangent modulus. But the tangent modulus does 



not measure the stiffness of the material when the material is 
resisting the working stress; this value, however, is frequently 
needed as, for example, in the design of reinforced concrete col¬ 
umns which involves the ratio of the moduli of elasticity of steel 
and concrete at the working stresses. The modulus of elasticity, 
therefore, is sometimes taken as the slope of a line OBC where the 
ordinate to B represents the working unit-stress; the slope of 
OBC is called the secant modulus. Although the secant modulus 
does not, according to the definition of stiffness, measure the 
stiffness of the material at the stress B'B, it is approximately an 














































DEFINITION AND MEASURE OF STIFFNESS 


357 


average value of the stiffness as the material is stressed up to the 
working stress. 

In Fig. 263 are shown the stress-strain curves for four struc¬ 
tural materials plotted to the same pair of axes and to the same 
scales. These curves show that at stresses below the elastic 
strengths of the materials, wood is only about one-twentieth as 
stiff as steel, and that cast iron is about one-half as stiff as 
steel. The medium carbon steel has the greatest static strength 




Fig. 262.—Methods of obtaining modulus of elasticity. 

and the low carbon steel is the most ductile, but both grades of 
steel have the same stiffness at stresses below the proportional 
limits. In fact the tensile moduli of elasticity of practically all 
steels including heat-treated alloy-steels are approximately the 
same; namely, 30,000,000 lb. per sq. in., although the elastic 
strengths may vary from 25,000 to 200,000 lb. per sq. in. The 
compressive modulus of elasticity is nearly the same as the ten¬ 
sile but the shearing modulus of elasticity of steel (and of most 
metals) is much less than the tensile or compressive modulus. In 
other words, the shearing strain in steel increases with shear¬ 
ing stress much faster than elongation increases with tensile stress. 

In plotting a stress-strain curve from test data it frequently 
happens that the curve does not pass through the origin; the 
slope of the curve, which represents the modulus of elasticity, is 










358 DEFINITIONS, MEASURING, MECHANICAL PROPERTIES 


not then expressed by the ratio -. To obtain the modulus of 

elasticity a line may be drawn through the origin and parallel 

to the plotted curve (Fig. 2626) and the value of ~ as obtained from 

this line will be the modulus of elasticity; or, the difference between 

two values of the unit-stress may be divided by the difference in 

( As 
i£ = —, Fig. 

2626). The scale of plotting should be such as to cause the curve 



Unit-strain 

Fig. 263.—Stress-strain diagrams for four structural materials. 

to make an angle of from 30° to 45° with the vertical in order to 
obtain a reliable value for the modulus. Further, the longer the 
gage length the more accurate will be the value of the modulus; 
a gage length of not less than 8 in. is recommended by the American 
Society for Testing Materials. The* extensometer used should 
measure the deformation on at least two sides of the specimen, 
and the same precautions concerning axial loads should be taken 
as in determining the proportional limit (see Art. 137a). Values 
of moduli of elasticity for various materials are given in the tables 
of Chapter XVI. 




































USE AND SIGNIFICANCE OF STIFFNESS 


359 


145. Use and Significance of Stiffness. —A distinction should 
be made between the stiffness of a structural member , such as an 
eye bar, a beam, or a shaft, etc., when subjected to loads and the 
stiffness of the material of which the member is made. The stiff¬ 
ness of the member depends on the size and form of the member 
as well as on the stiffness of the material. The manner in which 
the size and form of a member affects its stiffness is discussed in 
Part I and, as there pointed out, stiffness of a member rather than 
its strength may be the governing factor in a design. Under these 
conditions the stiffness of the material is an important property of 
the material. Thus a machine tool such as a planer lathe tool, 
drill press, or grinding machine may deflect too much for accurate 
work even though the stress in it is relatively small; a long shaft 
may twist so much that troublesome vibrations are set up under 
fluctuating loads even though the shaft is amply strong; a long 
floor beam may deflect so much that the plastered ceiling beneath 
will crack even though the maximum stress in the beam is well 
within the elastic strength of the material; the guides for a loco¬ 
motive cross-head may deflect enough to interfere with the running 
of the engine, etc. 

If the members in the above illustrations were made of steel 
the trouble could not be relieved by merely substituting a member 
made of a stronger grade of steel; for, if the loads and dimensions 
remain the same the stresses do not change, and hence the deflec¬ 
tion of the member could change only because of a change in the 
stiffness of the material, but the stiffness of the strongest (high 
carbon) steel is the same as that of the softest (low carbon) steel, 
consequently the deformation or deflection of the member would 
not change. 

If, however, the floor beam in the above illustration were made 
of wood, the beam could be made much stiffer by replacing it by 
one having the same dimensions but made of steel, for although 
the stress in the steel beam would be the same as that in the timber 
beam (the loads and dimensions remaining constant), the deflection 
of the steel beam would be much less. 

The property of stiffness is also of great importance in deter¬ 
mining the resistance of material to energy loads, that is, in deter¬ 
mining the amount of energy the material can absorb; for, as 
shown in Art. 116 and discussed in the following article, the amount 
of energy that a material can absorb per unit volume (cubic inch) 


360 DEFINITIONS, MEASURING, MECHANICAL PROPERTIES 


1 s ^ 

is - when stressed to its elastic strength, 

z hi 


Thus if the stiffness 


of a material is small (assuming the strength to remain constant) 
the capacity of the material to absorb energy without being stressed 
above its elastic strength is large. Thus the low stiffness of wood 
makes it desirable for railway ties and for spokes in automobile 
wheels, etc# 

Values of the moduli of elasticity of various materials are given 
in the tables of Chapter XVI. 


QUESTIONS 

29. Define stiffness. How is the stiffness of a material found? In what 
units is it expressed? 

30. Is the stiffness of structural steel when stressed above the yield-point 
greater or less than when stressed below the yield-point? 

31. How is the stiffness of a material measured if the material has a curved 
stress-strain diagram? Define tangent modulus and secant modulus. 

32. Compare the stiffness of steel, wood and cast-iron at stresses below 
their elastic strengths. 

33. Is the stiffness of steel at stresses below the proportional limit the 
same for all grades of steel? Compare- the tensile and shearing stiffness of 
steel at stresses below the proportional limits. 

34. Under what conditions is it desirable for a force-resisting member to be 
made of material having a high degree of stiffness? Having a low degree of 
stiffness? 


Resilience; Elastic Energy Strength 

146. Definition, Measure and Significance of Resilience.— 

Resilience is the property of a material that enables the material 
to give up or release energy (that is, to do work) as the stress is 
released. 

The energy recovered when the stress is released from the elastic 
limit is called elastic resilience , and the energy recovered per unit 
volume is called the modulus of elastic resilience of a material. 
Now all the work done in stressing a material to its elastic limit (or 
to any stress less than the elastic limit) is stored in the material 
and can be recovered as the stress is released; that is, none of 
the energy is dissipated in heat in causing structural damage, and 


RESILIENCE, ELASTIC ENERGY STRENGTH 


361 


hence, the work done per unit-volume in stressing the material 
to the elastic limit is equal numerically to the modulus of elastic 
resilience, k. Therefore, k is represented by the area OAB (Fig. 
264) and is equal to 

k 2 E’ 

as explained in Art. 116. 

But the elastic energy strength of a material is the amount of 
work that can be done on (or absorbed by) the material per unit 
volume in stressing the ma¬ 
terial to its elastic limit (see 
Art. 116); therefore, the mod¬ 
ulus of resilience is a measure 
of the elastic energy strength of 
the material. The elastic 
energy strength of a struc¬ 
tural member , however, de¬ 
pends on the form and di¬ 
mensions of the member as 
well as on the elastic energy 
strength of the material of 
which the member is made 
(see Chapter XIII). 


A 


^Hyper-elastic 



Resilience 


-Elastic Ky 

Resilience Y/y 


1 

§ 

E 


0 

Fig. 264. 


Unit-strain 

-Areas representing resilience. 

In obtaining the tensile modulus of resilience, s e and E in the 
above expression are the tensile elastic limit and the tensile modulus 
of elasticity, respectively; whereas, in expressing the shearing 
modulus of resilience s e and E are the shearing elastic limit and the 
shearing modulus of elasticity, respectively. 

The ideal material, then, for resisting energy loads that must 
not cause plastic deformation in the material is one having a high 
elastic strength and a low modulus of elasticity. Thus, the mod¬ 
ulus of resilience of high carbon and alloy steels is much larger 
than that of low carbon steel due solely to the greater elastic 
strength, since the modulus of elasticity is not influenced by the 
carbon or alloy content. Further, the shearing modulus of resil¬ 
ience of steel, k s , is approximately equal to the tensile modulus, k t . 
This is shown as follows: 

* 2 E, 

but from Art. 5 and 140, s s = 0.6s< and E s = %E t . 


and 


fc= 1 ^ 2 
‘ 2 E t ’ 






362 DEFINITIONS, MEASURING, MEACHINCAL PROPERTIES 


Therefore, 


. _l (0-6) 2 (, s.)i 2 
8 2 0.4 E t 


1 approximately. 

2 hi t 


Thus, the resistance that steel can offer to energy loads without 
acquiring plastic deformation is approximately independent of 
the kind of stress to which the steel is subjected (see Art. 113). 

On the other hand, bronze is as strong as some grades of steel 
and its stiffness (modulus of elasticity) is considerably lower. 
Therefore bronze may be nearly as good material for a spring, 
etc., as the poorer grades of spring steel. Likewise timber is low 
in stiffness and hence possesses a fair degree of resilience even 
though its elastic strength is low. (See Table XI of Chapter XVI 
for a comparison of the elastic energy strengths of various mate¬ 
rials) . 

If a bar is stressed beyond the yield-point of the material most 
of the work done is dissipated in causing structural damage. If, 
however, the stress is released from some value such as EC (Fig. 
264), the stress-strain diagram will take the form CD. That is, the 
part DE of the total unit-strain OE is elastic deformation, and the 
area DCE represents the energy recovered per unit-volume and 
hence measures the modulus of resilience corresponding to the 
stress EC. But it is convenient to use the term modulus of 
elastic resilience for the energy represented by the area OAB and 
to use the term modulus of hyper-elastic resilience for the energy 
recovered per unit volume when the stress in the material is 
released from a value above the elastic strength of the material. 

The modulus of hyper-elastic resilience is a property, however, 
of little use in selecting material for resisting loads, whereas the 
modulus of elastic resilience is of great importance in selecting 
materials for members to be subjected to energy loads that are to 
develop stresses less than the elastic strength of the material, such 
as springs, connecting rods of forging hammers and of rock drills, 
various automobile parts, etc. 

As shown in Table XI, the value of the modulus of elastic resil¬ 
ience of various grades of steel vary from about 15 to 670 in.-lb. 
per cu. in., the latter value being obtained from special alloy steel 
used for springs, axles, gears, etc., which must resist energy loads 
without being permanently distorted. 



TOUGHNESS; ULTIMATE ENERGY STRENGTH 


363 


QUESTIONS 

35 . Define resilience, elastic resilience, modulus of elastic resilience.. 
What is the quantitative expression for the modulus of elastic resilience? 

36 . Define elastic energy strength of a material and show that it is measured 
by the modulus of elastic resilience. 

37 . What use of material requires a high modulus of elastic resilience? 

38 . What is the modulus of hyper-elastic resilience? Is it of importance: 
in force-resisting members? 

39 . Compare the tensile and shearing moduli of elastic resilience of steel. 

Toughness; Ultimate Energy Strength 

147. Definition and Measure of Toughness.—Toughness is 
that property of a material that enables it to absorb energy while 
being stressed above its elastic strength, that is, while incurring 
plastic deformation. Thus, the more work that has to be done in 
stressing a material from its elastic limit to its ultimate strength 
the greater is the toughness of the material. 

Measure of Toughness .—One measure of the toughness of a 
material is the amount of work per unit-volume of the material 
required to rupture the 
material when subjected 
to a gradually increasing 
(static) load. This meas¬ 
ure of toughness is repre¬ 
sented by the area under | 
a stress-strain curve; this 
area, however, includes >§ 
the work done in stress¬ 
ing the material to the 
proportional limit (that 
is, the resilience of the 
material), but the area F 
under the straight-line 
part of the stress-strain 
diagram of ductile materials is negligible in comparison with the 
total area as is evident from an inspection of Fig. 250(a); and 
it can be neglected even for relatively brittle materials without. 



i. 265.—Toughness represented by areas 
under stress-strain curves. 










364 DEFINITIONS, MEASURING, MECHANICAL PROPERTIES 


serious error particularly since an exact quantitative measure of 
toughness is not as a rule required. 

The stress-strain diagrams in Fig. 265 show that the high carbon 
steel has the greatest static strength (both elastic and ultimate) 
and the greatest modulus of resilience (or elastic energy strength), 
that the medium carbon steel has the greatest toughness and that 
the low carbon steel has the greatest ductility. 

Since the area under the stress-strain diagram of a ductile 
material is roughly proportional to the product of the ultimate 
static strength and the percentage of elongation of the material, 
this product, sometimes called the toughness index number or merit 
number , is used frequently for comparative purposes as an approx¬ 
imate measure of toughness. (See Art. 116 for expressions from 
which an approximate value of the area under a stress-strain curve 
may be found.) Values for toughness of several materials are 
given in Table XI of Chapter XVI. 

148. Significance of Toughness.—When a static overload due 
to accident or other cause is applied to a member of a structure, 
resulting in stresses greater than the elastic strength of the mate¬ 
rial, the ultimate strength of the material determines whether the 
member will successfully resist the load or will rupture, that is, the 
reserve strength of the material (see Art. 138), if sufficient, becomes 
available in preventing rupture. On the other hand, if an energy 
overload, that is, an amount of energy in excess of the resilience 
of the member (see Art. 114), is applied to the member by a moving 
body, thereby causing stresses in the member greater than the 
elastic strength of the material, the toughness of the material 
determines whether the energy of the moving body will be 
absorbed by the member or will rupture the member. In other 
words, the toughness of a material is a measure of the ultimate 
energy strength of the material and it furnishes reserve strength 
for resisting excessive energy loads. 

In service, then, in which material is subjected to energy loads 
and in which the stresses are to be kept within the elastic strength 
of the material without provision for accidental overloads, etc., 
1 s<? 

the resilience, ^ -gr, of the material is of prime importance. Thus 

the material in piston rods on steam forging hammers and rock 
drills should have high resilience rather than toughness; likewise, 
the materials in most springs and many gears are selected because 


THE SINGLE-BLOW, NOTCHED-BAR IMPACT TEST 365 


of their high resilience. On the other hand, the material in the 
frame of a railway locomotive or car which in service is practically 
certain to be subjected to impact loads that will cause stresses 
above the yield-point of the material, toughness in addition to resili¬ 
ence is of importance. The same is true to a slightly lesser extent 
of the material in ships and bridges. For example, the stresses 
in the members of bridges under conditions of service are expected 
to be less than the static elastic strength of the material (except 
the localized stresses, see Art. 143) but if a minor derailment or 
other accident should occur on a bridge, requiring that a con¬ 
siderable amount of energy present in the moving train shall 
be absorbed by the bridge the material must possess toughness 
if the bridge does not collapse. Automobile frames, axles, gears, 
etc., furnish other examples of members for which toughness, in 
addition to resilience, is an essential property. 

149. The Single-blow Notched-bar Impact Test.—This test 
was devised in response to a demand for a quick and convenient 
method of determining the suitability of a material for resisting 
impact and energy loads. It was urged that the resistance of a 
material to impact should be determined from a test in which the 
load is applied by a moving body. It is doubtful, however, whether 
this test determines the ultimate impact or energy resistance 
(toughness) of a material as the material is used in engineering 
structures and machines , for reasons discussed below. 

The test is usually made with a Charpy (or Izod n ) pendulum 
machine (Fig. 266); the specimen (Fig. 267) is a notched rect¬ 
angular beam. The energy lost by the pendulum in rupturing 
the specimen is the value found from the test; this value is some¬ 
times referred to as the “ Charpy value.” 

In order to cause a specimen of ductile material to rupture 
when struck by the pendulum, instead of merely bending, the 
specimen must be notched. 12 The form of the notch has a marked 

11 The Izod machine is used mainly in France and England and is prac¬ 
tically the same as the Charpy machine except that the specimen used in the 
Izod machine is tested as a cantilever beam whereas the Charpy machine 
tests the specimen as a simple beam; the form of the notch also is usually 
somewhat different. 

12 Although the controlling idea in devising the test was that of applying 
the load with impact, it is now recognized that the significant factor in the 
test is the notched form of specimen and the accompanying concentration of 
stress at the root of the notch. 


366 DEFINITIONS, MEASURING, MECHANICAL PROPERTIES 


influence on the amount of energy required to rupture the speci¬ 
men and hence the values found from the tests are meaningless 
even for comparative purposes unless the specimen is standard¬ 
ized; two forms of specimens frequently used are shown in Fig. 
267. 

Now, the portion of the material in the specimen that absorbs 
the greater part of the energy is near the bottom of the notch but 



Fig. 266.—Single-blow, notched-bar impact testing machine. (The Charpy 
machine.) The pendulum when released from its position Z strikes the 
specimen 5 and rises to the position Z". Thus the center of gravity, 
G, of the pendulum lowers a distance (h-h") as the pointer N moves to 
the position N"; the pointer remains at N" after the pendulum swings 
back from its position Z" and hence, the angle of rise is indicated on 
the scale E. The energy expended by the pendulum in rupturing the 
specimen is calculated from the measured quantities and is the value 
obtained from the test. 

the amount of material involved is indefinite and indeterminate, 
and hence the toughness, or energy absorbed per unit volume, is 
indeterminate. Again, it is doubtful if the resistance of the mate¬ 
rial, as found in this test, is the resistance offered by a material 
to an impact or energy load as such loads occur in service; the 
high local stress at the root of the notch, where the section is 



























HARDNESS 


367 


abruptly reduced, requires that the material at the root of this 
notched section shall fail suddenly before other parts of the 
specimen can be stressed, and hence before those parts can be 
made to absorb much of the energy. This action produces sudden 
tearing of the material, and materials in machines and structures 
probably seldom if ever are subjected to such severe action, even 
though some machine members have rather abrupt changes of 
sections, such as at sections at roots of screw threads, tool marks 
and deep scratches, at roots of small radii, etc. (Some engineers, 
nevertheless, feel that the test is not too severe.) 

The test, however, is rather widely used and is of special value 
in determining whether certain heat treatments of steel have 
been carried out successfully since heat treatments that have been 
poorly done affect the value obtained from the single-blow notched- 
bar test very markedly. This fact is of .great importance since 
much of the material used in automobiles, aeroplanes, and many 
special machines are heat treated in order to secure high strength 
and light weight and they must also have considerable toughness 
or reserve impact strength to resist accidental impact overloads. 
By heat treating the proper steels these desirable properties can 
be produced provided the heat treatments are successfully carried 
out, and hence a convenient test that sharply differentiates between 
successful and poor heat treatments is of great value. But mate¬ 
rials should not be selected on the basis of Charpy values alone; 
the results of impact tests should be interpreted in connection with 
other properties of the material. 

QUESTIONS 

40 . Define toughness of a material. How is it measured? 

41 . What is meant by toughness index number or merit number? 

42 . What uses of material require a high degree of toughness? Give illus¬ 
trations. 

43 . Does the single-blow notched-bar impact test measure the toughness of 
the material? If not, why? What significance has this test? 

Hardness 

150. Definition and Measure of Hardness.—The meaning of 
hardness of a material is different with the different operations or 


368 DEFINITIONS, MEASURING, MECHANICAL PROPERTIES 


services in which the material is used; the operations in which 
hardness is of special importance being scratching, abrasion, 
cutting, and penetration or indentation. Hardness seldom if 
ever signifies a single physical property of a material, but it is, 
however, always associated with stresses that are accompanied 
by plastic or permanent deformations. Hardness, therefore, may 
be defined as the resistance which a material offers to the com¬ 
plex and indefinite stresses that are brought into action in operation 
involving scratching, abrasion, cutting and penetration. 

The particular hardness that is of greatest importance in con¬ 
nection with the material in load-resisting members is the resist¬ 
ance to penetration or indentation. Two methods of measuring 
the resistance of a material to indentation are in wide use; namely, 
the Brinell ball test, which is a slow or static penetration test, and 
the Shore scleroscope test, which is a rapid or dynamic penetration 
test. 

151. The Brinell Ball Test.—The Brinell machine (Fig. 268) 
applies a pressure of 3000 kilograms (or 500 kilograms for very 
soft material) for fifteen to thirty seconds to a ball 10 millimeters 
in diameter, and this ball, which rests on the surface of the test 
specimen, causes a permanent indentation in the specimen, the 
diameter or width of the indentation being measured by a simply 
and conveniently constructed microscope. The load in kilograms 
is divided by the surface of indentation in square millimeters and 
the resulting average intensity of pressure is called a Brinell num¬ 
ber. (Brinell numbers for various materials are given in Table I, 
Chapter XVI.) 

152. The Shore Scleroscope.—The Shore scleroscope (Fig. 
269), releases a small steel hammer from a height of about 10 
inches, allowing it to fall freely in a glass tube with a diameter 
slightly larger than that of the hammer. The steel hammer 
weighs 2 ounce and has a diamond striking-point the face of which 
is rounded to a definite radius. When the hammer strikes the 
specimen it penetrates the surface of the specimen before rebound¬ 
ing and hence produces a permanent but minute deformation. In 
so doing, part of the energy acquired by the hammer in falling is 
absorbed, and hence the height of rebound is less than the original 
height of fall. The rebound of the hammer is observed on the 
vertical scale and this reading (called the scleroscope number) 
is taken as the measure of the hardness of the material; the harder 


LIMITATIONS OF HARDNESS TESTS 


369 


the material the higher the rebound. The scale reading of 100 is 
frequently fixed as the height of rebound of the hammer when it 
strikes a special hardened steel surface; this surface is then called 
the standard surface. 



Fig. 268. —Brinell hardness testing machine. The ball B is pressed into the 
specimen S by the oil pressure on the piston L. The pressure is main¬ 
tained constant for about thirty seconds by pumping (by hand) so that 
the known weight W is kept in a floating position. The specimen rests 
on a spherical-seated bearing block. The diameter of the indentation 
indicates the hardness of the material. 

153. Limitations of the Tests.—The Brinell test is not satis¬ 
factory for testing extremely hard materials, since the ball itself 
deforms too much, nor is it satisfactory for thin sheets of material 
such as saw blades, etc. Further, in testing finished products 























































370 DEFINITIONS, MEASURING, MECHANICAL PROPERTIES 


whose surfaces must not be dented, the Brinell test cannot be 
used . 13 In these three respects the scleroscope is to be preferred, 
but, in general, the Brinell test is usually considered to be the more 
reliable, since in the Brinell test more of the material is tested and 



Fig. 269. —Scleroscope hardness-testing machine. The hammer W with its 
diamond D falls freely in the glass tube T. The diamond point strikes 
the specimen S and rebounds; the height of rebound is read on the 
scale E, the value read on the scale being called the “scleroscope hard¬ 
ness number.” By squeezing the rubber bulb R , the hammer is raised; 
the hammer is then held by the latch C until the bulb is again squeezed 
when the latch is released and the hammer falls. 

the character or condition of the material and of the surface affects 
the results less than in the scleroscope test. Thus rubber and wood 

13 A smaller machine called the “ Baby Brinell,” which uses a ball only 
ts inch in diameter, may be used with relatively thin material. 







































NEED AND SIGNIFICANCE OF HARDNESS 


371 


may give the same rebound with the scleroscope, although the two 
materials differ in hardness. 

154. Need and Significance of Hardness. —Since hardness is 
closely associated with the resisting stresses that accompany 
plastic deformation, the yield-point or ultimate strength gives a 
general index of the hardness of a material. In fact, test results 14 
show that there is a fairly definite linear relation between the ulti¬ 
mate tensile strength of carbon steel and the Brinell and sclero¬ 
scope numbers. Therefore, the Brinell and scleroscope tests form 
convenient methods of estimating the tensile strength of machined 
or other parts from which it is impossible or not feasible to cut a 
test specimen. 

R. R. Abbott found that the relations between the tensile 
ultimate strength, s u , and the Brinell number, B, and scleroscope 
number s, for carbon and alloy steels was expressed approximately 
by the following equations: 

s* = 0.70R - 26 = 4.0$ -15, 

and that the relation between the Brinell and scleroscope number 
was approximately 

£ = 5.5/S+28. 

But, as noted above, there is no single test that satisfactorily 
measures all of the various kinds of hardness required in service. 
Further, the degree of hardness desired in producing any object 
has been determined almost entirely by experience; the Brinell, 
scleroscope, or other hardness tests assist in attaining uniformity 
of product but not in determining the degree of hardness the mate¬ 
rial should have for the service required. 

Many structural and machine members require a relatively 
high degree of hardness in addition to some one or more other 
properties such as resilience, toughness, etc.; thus railway rails, 
locomotive and car wheel tires, automobile gears and axles, armor 
plate, cutting edges of steam shovels, jaws of stone crushers, dies 
for wire drawing, etc., must have high resistance to indentation 
or abrasion or both. And since strength and hardness of metals 
are, in general, dependent properties, steel used for the above- 

14 Abbott, R. R. Proceedings, A.S.T.M., Vol. 15, 1915, p. 43 (see also 
Jour, of Iron and Steel Inst. 1909). 


372 DEFINITIONS, MEASURING, MECHANICAL PROPERTIES 


mentioned parts is usually relatively high carbon steel since the 
strength and hardness of steel increase with the carbon content. 
But the ductility and toughness of steel decrease with the carbon 
content and hence for those members which require toughness in 
addition to hardness special alloy steels are used-which possess a 
fair degree of ductility and toughness as well as high strength 
and hardness. Thus automobile gears and axles are frequently 
made of chrome nickel steel, armor plate of chromium steel, jaws 
of crushers of chromium or manganese steel, etc. And as noted 
above, after experience has demonstrated the kind of material to 
be used for a given service and the treatment to be given the 
material, a hardness test is frequently a convenient way of deter¬ 
mining whether the desired properties are being obtained. 

Resistance to Wear .—Sometimes the property of hardness as 
defined above is made to include “ resistance to wear.” However, 
resistance to wear depends on the ductility (or plasticity) of the 
material perhaps more than on its strength-hardness. That is, a 
good wearing metal must resist the displacement of its particles 
(and hence have hardness) but when they are displaced it must 
resist the removal of the particles from the body (and hence have 
plasticity). In general, if two metals have the same tensile 
strength the one having the greater ductility will possess the greater 
resistance to wear, although the resistance to wear of both metals 
would be increased by increasing the tensile strengths and the 
fineness of crystalline structure, assuming the ductility to remain 
constant. 


QUESTIONS 

44 . Define hardness of a material. 

45 . Describe the Brinell ball test and the Shore scleroscope test. What 
are the advantages and disadvantages of each? 

46 . Can either of these tests be used to determine the hardness needed for 
a given use? If not, how is this determined? 

47 . Give examples of uses of material in which hardness is one of the essen¬ 
tial properties of the material. 


Fatigue Strength 

155. Definition and Measure of Fatigue Strength. —The 

fatigue strength of a material is the greatest resistance the mate- 


FATIGUE STRENGTH 


373 


rial can offer, without rupturing, when subjected to a load that is 
repeated a great many times. As discussed in Chapter XIV, 
steel may rupture when subjected to a repeated load even when the 
calculated stress in the material is considerably less than the 
static proportional limit or elastic limit of the material. In other 
words, the fatigue ultimate strength of the material may be not 
only less than the static ultimate strength but even less than the 
static elastic strength of the material. Further, a member that 
fails due to a repeated load seldom gives warning of the approach¬ 
ing rupture; that is, no measurable plastic yielding of the member 
occurs at a stress less than the stress at which rupture occurs. 
Thus, there is no measurable elastic fatigue strength as there is 
with static strength; the only measurable fatigue strength is the 
ultimate or rupture fatigue strength. (See Art. 126 for an explan¬ 
ation of a fatigue failure.) 

The measure of the fatigue (ultimate) strength of a material 
is the greatest unit-stress in the material that can be repeated an 
indefinitely large number of times without causing the material 
to rupture. This unit-stress is called the endurance limit of the 
material (see Art. 125). Experimental investigations 15 have 
established, so far, endurance limits only for wrought ferrous 
metals (rolled and forged steel and wrought iron); comparatively 
little work has been done on cast-iron or steel castings or on non- 
ferrous metals. 

Values of the endurance limits (fatigue strengths) of various 
grades of iron and steel with completely reversed bending stress 
are given in Table 7 of Chaper XIV; and in Table XII of Chapter 
XVI; and the use of these values in obtaining the endurance 
limits with direct axial stress, with shearing stress and with various 
ranges of stress are discussed in Chapter XIV. The main object 
of this section is to discuss various methods of determining the 
endurance limit of materials. 

156. Methods of Determining the Endurance Limit. —(a) 

Direct Method. The direct method of determining the endurance 
limit of a material is to test several specimens of the material 
as described in Art. 125, and to plot an s-iV diagram (see Figs. 
237 and 238). Repeated stress tests have not yet been standard- 

15 See Bulletins 124, 136 and 142 of the Engineering Experiment Station of 
the University of Illinois; an extensive bibliography is given in Bulletins 
124 and 142. 


374 DEFINITIONS, MEASURING, MECHANICAL PROPERTIES 


ized, but two types of repeated stress testing machines eommonly 
used for determining the endurance limit will here be described 
briefly. 

In Fig. 270 is shown a machine which applies repeated flexural 
stress by means of a crank and connecting rod and measures the 
bending moment applied to the specimen, by means of the com¬ 
pression of calibrated 
springs. Power is 
furnished by a motor 
M (or from a line 
shaft) and a crank C 
with an adjustable 
throw is driven by the 
motor. The crank is 
attached to a connect¬ 
ing rod R which bends 
a specimen S back and 
forth. The motion of 
the specimen is resist¬ 
ed by springs G acting 
through a bent lever 
A. The amount of 
bending moment ap¬ 
plied to the specimen 
may be varied by 
changing the throw of 
the crank and is meas¬ 
ured by the amount 
of compression of the 
springs G. This compression is indicated by the throw of the 
arm I to the end of which is attached a pencil that records 
the throw on paper wrapped round the drum D. The drum 
D is rotated by a worm and wheel drive from the main shaft 
of the machine, and there is consequently recorded on the paper 
a diagram whose width is a measure of the bending moment 
on the specimen and whose length is a measure of the number 
of applications of stress to the specimen. The number of 
applications of stress is also indicated by a counter K. From the 
bending moment the maximum unit-stress applied to the specimen 
can be determined. Usually this type of machine is used to pro- 



Fig. 270.—Upton-Lewis repeated-stress testing 
machine. 







































DETERMINING THE ENDURANCE LIMIT 


375 


duce reversals of bending stress, but by varying the springs G 
other stress ranges can be applied to the specimen. The Upton- 
Lewis machine (Fig. 270) is the commonest example of this type 
of machine used in the United States. 

Fig. 271 shows in diagram a testing machine which produces 
reversals of bending stress by the use of a rotating flexure specimen. 
The specimen S is supported on ball bearings B and driven by a 
pulley P. Weights are hung from a second set of ball bearings B' } 
and these weights cause bending stresses in the specimen. The 
bending stress in the upper fibers of the specimen is compression, 
and in the lower fibers, tension. As the specimen is rotated the 



stress for any fiber changes from compression to tension, and the 
stress is completely reversed. The maximum unit-stress for both 
tension and compression can be computed from the amount of 
weights W applied, the dimensions of the specimen, and the dis¬ 
tances between bearings; the bending moment is constant for all 
sections between the two center bearings. A counter K indicates 
the number of reversals of stress given to the specimen, and when 
the specimen breaks the counter automatically stops. 

(6) Rise-of -Temperature Test .—The determination of the endur¬ 
ance limit by the actual application of repeated stress as outlined 
above requires so much time that it is not as serviceable in com¬ 
mercial testing as is desired, and hence various short-time or accel¬ 
erated tests for determining the endurance limit have been tried; 
one such test that has proven to be fairly reliable for wrought 
























































376 DEFINITIONS, MEASURING, MECHANICAL PROPERTIES 


ferrous metals is the rise-of-temperature test. 16 This test consists 
in subjecting a specimen to a small number of reversals (about 500) 
of a known stress and measuring the rise in temperature of the 
specimen, at the section subjected to the maximum stress, by 
means of a thermo-couple and a galvanometer, or by some other 
method; the same test is then repeated two or three times, increas¬ 
ing the stress in each successive test until a well-marked rise of 
temperature can be detected. The endurance limit is taken from 
a stress-temperature diagram as the unit-stress at which the rate 
of rise of temperature shows a marked increase. The values of the 
endurance limit for wrought ferrous metals found by this method 
vary but little from those found by the direct long-time repeated- 
stress tests. However, this test, or any other short-time test thus 
far developed, is essentially supplementary to, rather than a sub¬ 
stitute for, the long-time endurance test. 

(c) Relation of Endurance Limit to Other Physical Properties .— 
It was formerly thought that the elastic limit (or proportional 
limit) was a measure of the resistance of a material to repeated 
stress. Now, as pointed out in Art. 137, the elastic limit is an 
index of the beginning of appreciable plastic action in a material 
although it is rather an uncertain quantity as discussed in Art. 137, 
whereas the failure of a material due to repeated stress, as shown 
by microscope examination of specimens, seems to be a progressive 
tearing apart, or shearing apart (rupturing) across minute areas of 
the metal (see Art. 126). Evidence of the truth of this statement 
is found by examining the photomicrographs in Figs. 272 (6) and (c) 
which show the surface of a piece of steel stressed slightly beyond 
the yield-point when subjected to a single steady load. In Fig. 
272(6) may be seen a few slip lines, which, under repeated loading 9 
might develop into a fatigue failure similar to that shown in Fig. 
273(6); there may also be seen in Fig. 272(6) dark patches which 
locate “ valleys ” caused by the plastic wrinkling of the surface 
of the steel; this wrinkling is more pronounced in Fig. 272(c). 
The beginning of this plastic action coincides approximately with 
the proportional limit of the steel rather than with the ultimate 
strength, and spread of the wrinkled areas is in contrast with the 
gradual spread of minute fractures which constitutes fatigue failure 
as indicated in Fig. 273(6). Thus it does not appear strange that 

16 See Bulletin 124 of the Engineering Experiment Station of the Univer¬ 
sity of Illinois. 


DETERMINING THE ENDURANCE LIMIT 


377 


the results of repeated stress tests show a fairly good correlation 
between the endurance limit and the ultimate tensile strength of 
iron or steel, whereas the correlation between the proportional 
limit (or yield-point) and the endurance limit is not so close. 



c 


Fig. 272.—Photomicrographs of steel subjected to a static load: (a) Speci¬ 
men in unstressed state. (6) Specimen stressed slightly beyond the 
yield-point, (c) Specimen stressed considerably beyond the yield-point. 

Experimental data (see Table XII of Chapter XVI) are rather 
consistent in showing that, for a rather wide range of steels, the 
endurance limit with completely reversed bending stress is approx¬ 
imately 0.5 of the tensile ultimate stress 17 (nearly all the steels 

17 As pointed out in Chapter XIV, the endurance limit with complete 
reversals of axial stress (tension-compression) may be as low as 0.60 of the 
endurance limit with reversal of bending stress. With reversed shearing stress 
the endurance limit is approximately 0.55 of the endurance limit with reversed 
bending stress. 



378 DEFINITIONS, MEASURING, MECHANICAL PROPERTIES 

tested give values between 0.4 and 0.6). If actual test data from 
repeated-stress tests are available the endurance limit should be 
determined from such test data, but if such data are not available 
it is recommended that the endurance limit for a wrought ferrous 
metal with completely reversed bending stress be estimated to be 
0.45 per cent of the static ultimate tensile strength. Further, since 
the “ Brinell hardness number ” (Art. 151) is a rough index of the 
tensile ultimate strength it is also a rough index of the fatigue 
strength of the material. 



Fig. 273.—Photomicrograph of steel subjected to a repeated load, (a) 
Specimen in unstressed state. (6) Specimen after being subjected to 
many repetitions of stress considerably above the endurance limit. 


No values can be given for estimating the fatigue strength of 
non-ferrous metals or of non-metallic materials until more test 
data are available. 


QUESTIONS 

48 . Define fatigue strength of a material. What is the measure of fatigue 
strength? 

49 . Is there any experimental evidence of an elastic fatigue strength? 

50 . Describe three methods of obtaining the endurance limit of a material 
from tests. 

51 . What evidence is there that the static ultimate strength is a better 
criterion of the fatigues trength of a material than is the static elastic limit or 
yield-point? 


CHAPTER XVI 

TABLES OF PROPERTIES OF MATERIALS 

By H. F. Moore 1 and F. B. Seely 

157. Use of Tables. —In this chapter are grouped, for con¬ 
venience, tables giving average values of the properties of the more 
common structural materials. In using these tables it should be 
remembered that the values given are average values, and that the 
values of the properties of any one material may vary considerably 
from those given in these tables. Therefore, the values given 
should not be used blindly; the uniformity and general quality 
of the material should be considered in addition to the properties. 
If possible, structures and machines in which the material has been 
used should be studied. When average values of the properties 
of a material are the only information available the values used in 
design should err on the side of safety. 

1 Research Professor of Engineering Materials, University of Illinois. 


379 


380 


TABLES OF PROPERTIES OF MATERIALS 


^-li§I§2§SiSiIilISS ii i : 


Elonga¬ 
tion in 2 

Inches 
Per Cent 

Slight 

Slight 

10.0 

48.0 

35.0 

45.0 

35.0 

18.0 

30.0 

25.0 

25.0 

15.0 

15.0 

15.0 

10.0 

9.0 

2.0 

10.0 

9.0 

1.0 

25.0 

9.0 

5.0 

14.0 

Slight 

(c) 

(c) 

Modulus of Elasticity 

Lb. per Sq. In. 

1 

6,000,000 

(c) 

10,500.000 

12,000,000 

10,000,000 

12,000,000 

12,000,000 

12,000,000 

12,000,000 

12,000,000 

12,000,000 

12,000,000 

12,000,000 

12,000,000 

12,000,000 

12,000,000 

12,000,000 

12,000,000 

12,000,000 

12,000,000 

12,000,000 

12,000,000 

12,000,000 

12,000,000 

(C) 

(/) 

(.0 

Tension 

15,000,000 
(c) 

23,000,000 

30,000,000 

27,000,000 

30,000,000 

30,000,000 

30,000,000 

30,000,000 

30,000,000 

30,000,000 

30,000,000 

30,000,000 

30,000,000 

30,000,000 

30,000,000 

30,000,000 

30,000,000 

30,000,000 

30,000,000 

30,000,000 

30,000,000 

30,000,000 

30,000,000 

(c) 

30,000,000 

30,000,000 

Strength in Shear 

Lb. per Sq. In. 

j 

§ 3 's. 

3- 

Proportional 

Elastic 

Limit 

ss i!iiiiiii!!Siiiii!i n i 3 

s’s’ s’ - 

Strength in Compression 
Lb. per Sq. In. 

Ultimate 

II 

jooggggggggggggggggggg go 22 SS 

Proportional 

Elastic 

Limit 

iiiiiiiiiiiiiiiiiiiii n i« ,, 

S’S SS 

Strength In Tension 
Lb- per Sq. In. 

Ultimate 

iiiiiiiiiiiiiiiiiiiii n §§ n 

s’s’s'3’s’s’s’s's'es’g’8’s’|s’|g’s’s’s g’S gg SS 

Proportional 

Elastic 

Limit 

iiiiiiiiiiiiiiiiiiiii n i§ n 

S 2’s’2’s’s’s’s’s’s's’s5's's’g’s’s’§s’gs S’S’ gg g’S 












































TABLES OF PROPERTIES OF MATERIALS 


381 


TABLE II 

Average Values for Static Strength and Ductility of Various 
Non-ferrous Metals and Alloys 

Values given are based on test data from various testing laboratories. 


Metal or Alloy 

Approximate 

Composition 

Per Cent 

Weight 
Lb. per 
Cu. In. 

Strength in 1 

Proportional 

Elastic 

Limit 

Lb. per 

Sq. In. 

'’ension (a) 

Ultimate 
Tensile 
Strength 
Lb. per 
Sq. In. 

Elonga¬ 
tion in 

2 In. 
Per Cent 

Copper, annealed. 

Copper, 100 

0.320 

3,200 

32,000 

56 

cold-drawn... 




38,000 

50,000 

8 

Zinc, cast. 

Zine. 100 

0.253 


9,000 

Slight 

rolled. 




5,000 

24,000 

50 

Tin, cast. 

Tin. inn 

0.260 


4,000 (6) 


rolled. 





5,000 (c) 


Read, east . 

Lead, inn 

0.410 


1,700 (d) 


rolled. 





3,300 (e) 


Common brass, cast... 

J 

Copper, 60; 

0.290 

20,000 

55,000 

20 

rolled. 

1 

zinc, 40 


25,000 

65,000 

30 

Phosphor bronze, 







cast. 


Copper, 95; 

0.32 

10,000 

32,000 

7 

rolled. 


tin, 4.9; 


40,000 

65,000 

30 

hard-drawn.... 


phosphorus 



105,000 

5 

spring wire.... 


trace 





Aluminum bronze, 







cast. 


Copper, 90; 

0.27 

25,000 

60,000 

25 

rolled. 


aluminum, 10 


30,000 

70,000 

30 



Nickel, 67: 





Monel metal, 


copper, 28; 

0.32 




cast. 


iron + carbon 


37,000 

72,000 (/) 

34 

rolled. 


+manganese 


50,000 

85,000 

42 



-(-silicon, 5 






General Note: A large number of special alloys of copper, tin, zinc, and other metals 
are in use whose physical properties differ somewhat from those tabulated here. The 
physical properties tabulated above are only general averages, and they may be materially 
modified by heat treatment, and very markedly modified by cold-rolling or cold drawing. 

(a) Where no values for compressive strength are noted the ultimate in compression 
may be assumed as having the same value as the proportional limit in tension. The strength 
in shear may be taken as 60 per cent of the strength in tension. 

( b) Ultimate in compression, 6400 lb. per sq. in. 

(c) Modulus of elasticity in tension, 4,000,000 lb. per sq. in. 

( d ) Modulus of elasticity in tension, 700,000 lb. per sq. in. 

( e) Modulus of elasticity in tension, 1,000,000 lb. per sq. in. 

(/) Modulus of elasticity in tension, 22,000,000 lb. per sq. in. 


































382 


TABLES OF PROPERTIES OF MATERIALS 


TABLE III 

Average Values for Static Strength and Ductility of Light 

Metal Alloys 

The values given in this table are based on data from various testing labora¬ 
tories, especially from the laboratory of the U. S. Air Service at McCook 
Field, Dayton, Ohio. 


Metal 

Weight 
Lb. per 
Cu. In. 

Tensile strength* 

Lb. per Sq. In. 

Elongation 
In 2 In. 
Per Cent 

Proportional 
Elastic Limit 

Ultimate 

Commercial aluminum, 99% pure: 

cast. 

0.093 

9,000 

13,000 

20 

rolled and annealed. 

0.097 

8,500 

13,500 

23 

hard-drawn. 

0.097 

20,000 

30,000 

4 

hard-drawn wire. 

0.097 

30,000 

40,000 

4 

Aluminum 96%, copper 4%: 

cast. 

0.104 

11,500 

19,500 

12 

hard-drawn. 

0.104 

35,000 

41,000 

5 

Duralumin; aluminum 96%, magnesium 1%, 
copper 2.9%, traces of iron, silicon, and 
manganese: 

annealed. 

0.102 

6,800 

25,200 

18 

tempered. 

0.102 

18,500 

51,200 

29 

Electron metal, magnesium 95%, zinc 4.4%, 
small quantities of copper, iron, silicon, 
and aluminum: 

rolled. 

0.065 

6,800 

25,200 

25 


* The compressive strength of metals here tabulated may be safely taken as equal to 
the proportional elastic limit in tension. The strength in shear may safely be taken as 60 
per cent of the tensile strength. 























TABLES OF PROPERTIES OF MATERIALS 


383 


TABLE IV 

General Properties and Uses of Wood 


Wood 

General Properties 

Used for 

Hard Woods: 



White Oak. 

Strong and tough, close grained, 
splits with difficulty, heavy, 
must be carefully seasoned to 
avoid checking. 

Ties, vehicle and furniture 
making, interior finish, fram¬ 
ing where great strength is of 
first consequence. 

Red Oak. 

A softer, weaker, and more por¬ 
ous kind of oak than white oak. 

Ties, interior finish, furniture. 

Hickory. 

The strongest, toughest, and 
heaviest of American woods. 
Susceptible to the attacks of 
boring insects. 

Vehicle and agricultural imple¬ 
ment manufacture. 

Maple. 

Rather coarse grained, but heavy 
and strong. It takes a fine 
polish. 

Flooring, furniture, interior 
finish. 

Elm. 

Strong and tough, but difficult to 
split and shape. It warps 
badly. 

Vehicle and ship building. 

Ash. 

Strong, but not very tough. 

Interior finish. 

Soft Woods: 



Spruce. 

Light, soft, straight grained, re¬ 
sistant to decay. 

Framing timbers, piles, under 
water construction. 

Douglas Fir. 

(Oregon Fir) 

Strong, though rather variable in 
quality, durable. 

All kinds of construction. 

White Pine. 

Light, soft, straight grained, not 
very strong. 

Pattern making, interior finish. 

Norway Pine. 

Hard, light, coarse grained. 

All kinds of construction. 

(Red Pine) 



Western Pine. 

Trade name for a number of kinds 
of wood with general character¬ 
istics somewhat like Norway 
pine. 

General construction work. 

Yellow Pine. 

The strongest and toughest of the 
soft woods. Heavy, decays in 
contact with soil. 

Heavy framing timbers, flooring, 

Hemlock. 

Light, brittle, splits easily. 

Cheap framing lumber, crates 
and boxes. 

Tamarack (Larch).. 

Strong, heavy, durable. 

Ties, sills, posts and poles, ship 
lumber. 

Cedar. 

Light, durable, but not strong. 

Water tanks, shingles, posts, 
fencing, boats. 

Redwood. 

Durable in contact with soil 
Weak and brittle, but very 
easy to work. 

Ties, posts and poles, general 
construction. 

Cypress. 

Very durable, light, close grained 

Shingles, poles, siding, interior 

easily worked, takes high polish 

finish, general building lumber. 






























Static Strength and Stiffness of Wood 

The values given in this table are average values based on results of a large number of tests mainly by the U. S. Forest Service 
laboratories. 


384 


TABLES OF PROPERTIES OF MATERIALS 


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(а) For most small pieces of wood it is possible to use air-dry wood and to keep it in that condition. For cases where there is likelihood of reab¬ 
sorption of water the values given will be diminished by about one-third. For the rare cases in which kiln-dry wood can be used and protected from 
reabsorbing water the values given will be increased by about one-third. 

(б) The wood marketed as Western pine includes several kinds of wood. It has about the same average strength values as Norway pine. 





































































TABLES OF PROPERTIES OF MATERIALS 


385 


TABLE VI 

Compressive Strength of Piers 

The values in this table are based on test data from Watertown Arsenal, 
the U. S. Bureau of Standards, Cornell University, and the University of 
Illinois. 


Make-up of Pier 

Ultimate 

Compressive 

Strength 

Lb. per Sq. In. 

Terra-cotta block, Portland cement mortar. 

3000 

Vitrified brick, Portland cement mortar. 

2800 

Pressed brick, Portland cement mortar. 

2000 

Pressed brick, lime mortar. 

1400 

Common brick, Portland cement mortar. 

1000 

Common brick, lime mortar. 

700 

Sand-lime brick, lime mortar. 

500* 


* Strength of sand-lime brick piers is estimated from comparative strength of individual 
sand-lime bricks and common bricks. 


TABLE VII 


Compressive Strength of Portland Cement Concrete 


The values in this table are from the report of the Joint Committee on 
Concrete and Reinforced Concrete. The values are based on test data from 
specimens in the form of cylinders 8 inches diameter by 16 inches long, stored 
under laboratory conditions and tested when 28 days old. The fine aggregate 
except for gravel and cinders is sand, and enough was used to fill the voids in 
the coarse aggregate; usually the ratio (by volume) of fine aggregate to coarse 
aggregate was about 1:2. 



Granite, trap rock. 

3300 

2800 

2200 

1800 

Gravel, hard limestone, hard sandstone. 

3000 

2500 

2200 

1800 

Soft limestone, soft sandstone. 

2200 

1800 

1500 

1200 

Cinders. 

800 

700 

600 

500 


1400 

1400 

1000 

400 
































386 


TABLES OF PROPERTIES OF MATERIALS 


TABLE VIII 

Results of Shearing Tests of Portland Cement Concrete 

The values in this table are summaries of results given in Bulletin No. 8 of 
the University of Illinois Engineering Experiment Station. It is probable 
that concrete, a brittle material, fails in tension on inclined planes rather than 
in shear. The values here given are the unit-stresses in shear at failure for 
specimens so tested that flexure, direct tension, and direct compression were 
minimized. The concrete was 60 days old when tested. The aggregate used 
was bank sand and soft limestone. 



Ultimate, Lb. per Sq. In. 

Proportion of 
Cement to 
Aggregate 

Computed 
Unit-stress 
in Shear at 

Compressive 
Strength 
as Given by 


failure; 

Compression 


Shearing Tests 

Tests 

1 : 6 

1290 

2430 

1 : 9 

1090 

1290 


TABLE IX 

Static Strength and Stiffness of American Building Stone 
Values based on test data from Watertown Arsenal. 


Stone 

Weight 
(Av.) 
Lb. per 
Cu. Ft. 

Ultimate in Compres¬ 
sion . 

Lb. per Sq. In. 

Computed 
Ultimate 
in Shear 
Lb. per 
Sq. In. Av. 

Modulus of Rupture 
(Flexure) 

Lb. per Sq. In. 

Modulus of 
Elasticity 
(Flexure) 
Lb. per 
Sq. In. 

Max. 

Min. 

Av. 

Max. 

Min. 

Av. 

Granite 

165 

26,000 

15,000 

20,000 

2300 

2200 

1200 

1600 

7,500,000 

Marble. 

170 

16,100 

10,300 

12,500 

1300 

2300 

800 

1500 

8,200,000 

Limestone. . 

160 

20,000 

3,200 

9,000 

1400 

2700 

250 

1200 

8,400,000 

Sandstone. . 

135 

19,000 

6,700 

12,500 

1700 

2200 

500 

1500 

3,300,000 

Slate . 

175 



15,000 




8500 

14,000,000 



































TABLES OF PROPERTIES OF MATERIALS 


387 


TABLE X 

Working Stresses for Structural Materials Subjected to 
Static Loading 

The values in this table are from the building ordinances of the city of 
Chicago. 

Allowable Compressive Stresses for Masonry (lb. per sq. in.): 


Coursed rubble, Portland cement mortar. 200 

Ordinary rubble, Portland cement mortar. 100 

Coursed rubble, lime mortar. 120 

Ordinary rubble, lime mortar. 60 

First-class granite masonry, Portland cement mortar. 600 

First-class limestone and sandstone masonry, Portland 

cement mortar. 400 

Portland cement concrete, 1-2-4 mixture, machine mixed. 400 

Portland cement concrete, l-2£-5 mixture, machine mixed.... 350 

Portland cement concrete, 1-3-6 mixture, machine mixed. 300 

Paving brick masonry, Portland cement mortar. 350 

Selected hard common brick masonry, Portland cement 

mortar. 200 

Common brick masonry, Portland cement mortar. 175 

Common brick masonry, lime mortar. 100 


Allowable Stresses for Timber (lb. per sq. in.): 


Wood 

Extreme 

Fibers 

in 

Beams 

Compression 
Along the 
Grain* 

Compression 
Across the 
Grain 

Shear 

Along 

the 

Grain 

Douglas fir and long leaf yellow 
pine. 

1300 

1100 

250 

130 

Oak. 

1200 

900 

500 

200 

Short leaf yellow pine. 

1000 

800 

250 

120 

Norway pine and white pine. . . 

800 

700 

200 

80 

Hemlock. 

600 

500 

150 

60 


* For columns these values are too high; a column formula must be used. 

























388 


TABLES OF PROPERTIES OF MATERIALS 


Allowable Stresses for Iron and Steel (lb. per sq. in.): 


Kind of Stress 

Rolled 

Steel 

Steel 

Castings 

Wrought 

Iron 

Cast 

Iron 

Tension on net section. 

16,000 

16,000 

12,000 


Max. compression on gross section. 

14,000 

14,000 

10,000 

10,000 

Bending on extreme fiber. 

16,000 

16,000 

12,000 


Bending on extreme fiber, tension. 

3,000 

Bending on extreme fiber, compression 




10,000 

Bending on extreme fiber pins . 

25,000 




Shear, pins and shop-driven rivets. ... 

12,000 




Shear, field-driven rivets 

10,000 




Shear on rolled steel shapes 

12,000 




Shear plate girder webs, gross section.. 

10,000 




Shear on brackets. 



2,000 

Bearing, pins and shop-driven rivets. . 

25,000 



Bearing, field-driven rivets. 

20,000 









TABLE XI 

Average Values of Modulus of Resilience (or Elastic Energy 
Strength) and Toughness (or Ultimate Energy Strength) 


Material 

Tensile 
Propor¬ 
tional 
Limit 
(lb. per 
sq. in.) 

s e 

Tensile 
Ultimate 
Strength 
(lb. per 
sq. in.) 

Su 

Tensile 
Modulus 
of Elastic¬ 
ity 

(lb. per 
sq. in.) 

E 

Ultimate 
Elonga¬ 
tion Per 
Inch of 
Length 
(in.) 
e w 

Tensile 
Modulus 
of Resili¬ 
ence (in.-lb. 
per cu. in.) 

1 s e 2 

2 E 

Toughness 
in Tension 
(Represented 
by area under 
Stress-strain 
Diagram) 
(in.-lb. per 
cu. in.) 

Low carbon steel. 

30,000 

60,000 

30,000,000 

0.35 

15.0 

15,700 

Medium carbon steel.. 

45,000 

85,000 

30,000,000 

0.25 

33.7 

16,300 

High carbon steel. 

Special alloy steel 

75,000 

120,000 

30,000,000 

0.08 

94.0 

5,100 

(Heat treated). 

200,000 

230,000 

30,000,000 

0.12 

667.0 

22,000 

Gray cast iron. 

6,000 

20,000 

15,000,000 

0.005 

1.2 

70 

Malleable cast iron. .. 

20,000 

50,000 

23,000,000 

0.10 

17.4 

3,800 

Rolled bronze. 

40,000 

65,000 

14,000,000 

0.30 

57.2 

15,500 

Timber (Hickory).... 

5,500* 

10,000* 

2,400,000* 


6.32* 



* In compression. 










































TABLES OF PROPERTIES OF MATERIALS 


389 


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APPENDIX I 


FIRST MOMENTS AND CENTROIDS OF AREAS 

158. Definitions.—The moment of an area with respect to an 
axis is the algebraic sum of the moments of the elementary parts of 
the area, the moment of each part being the product of the ele¬ 
mentary area and the perpendicular distance from the elementary 
area to the axis. This moment is called the first moment to dis¬ 
tinguish it from the moment 
of inertia (or second moment, 
see Appendix II) of the area. 

The first moment of an area 
is sometimes called the statical 
moment of the area. 

The centroid of an area is 
a point whose distance (called 
a centroidal distance) from 
any axis times the total area 
is equal to the moment of the 
area with respect to that axis. 

Hence, the coordinates, x and 
y , of the centroid C of an area 
the equations 

ax = ^xda and ay=Jyda .(242) 

Or, expressed in another way, the centroid of an area is that 
point at which the whole area may be conceived to be concen¬ 
trated and have the same moment with respect to any axis as has 
the actual (distributed) area. 

If an area is symmetrical with respect to an axis, the centroid 
of the area lies in the given axis. This statement is evident from 

391 














392 


APPENDIX I 


the fact that the moments of the areas on the opposite sides of the 
axis are numerically equal but of opposite sign. If an area is 
symmetrical with respect to each of two axes, the centroid of the 
area is the point of intersection of the two axes. 

159. Centroids Found by Integration.—In determining the 
centroid of an area by the method of integration, from the equa¬ 
tions in the preceding article, it is possible to select the element 
of area in various ways and to express the element in terms of either 
Cartesian or polar coordinates. The resulting integral may be a 
single or a double integral, depending on the way the element is 
selected. The integral, of course, is a definite integral, the limits 
of integration depending on the boundary curve of the area. In 
any case the element of area must be taken so that: 

1. All points of the element are the same distance from the 
line about which moments are taken; otherwise, the distance from 
the line to the element will be indefinite. 

2. The centroid of the element is known, in which case the 
moment of the element about the moment axis is the product of 
the element and the distance of its centroid from the axis or plane. 

The centroids of some of the common areas will be found in the 
following illustrative problems: 


ILLUSTRATIVE PROBLEMS 

Find, by the method of integration, the centroids of the following areas 
with respect to the axes indicated. 

Problem 246. Area of a Triangle.—In accordance with the second of the 

above rules the elements of area will be 
taken as strips parallel to the base of the 
triangle (Fig. 275). Since each element 
is bisected by the medium drawn from the 
vertex opposite the base, the centroid of 
each element, and hence of the entire 
area, lies on this median. If x denotes the 
width of the strip, the area of the strip is 
da — xdy. Thus. 



ay 


-f 


xydy. 


From similar triangles, the relation 
between x and y is, 


x b ^ 

;-=- or x = - (h—y). 

h—y h h 










FIRST MOMENTS AND CENTROIDS OF AREAS 393 


Hence, 



Therefore, 


The centroid of a triangular area, then, is on the median line at a distance of 
one-third of the altitude from the base. 

Problem 247. Sector of a Circle. —First Method .—The element of area 
will be selected in accordance with the first of the above rules as indicated in 
Fig. 276. Since the area is symmetrical with respect to the z-axis, the cen¬ 
troid lies on this axis, and hence y — 0. The value of x may then be found 
from the equation, 


a xda 




p cos 6- pdpdd = lr 3 sin a. 


Therefore, 


fr 3 sin a 


fr 3 sin a 2 r sin a 


a 


r 2 a 3 a 




Fig. 276. 


Fig. 277. 


Second Method— In accordance with the second of the above rules, the 
element of area will be selected as a triangle, as indicated in Fig. 278. The 
area of the triangle is hr 2 dd and the distance of its centroid from the y-axis is 












394 


APPENDIX I 


f r cos 8. Hence, the moment of the triangle with respect to the ?/-axis is 
cos 6d8, and x is obtained from the equation, 



cos ddd 


Therefore, 


= fr 3 sin a. 

__ fr 3 sin a 2 r sin a 

r 2 <x 3 a 



Problem 248. Area of Quadrant of an Ellipse.—The semi-axes of the ellipse 
will be denoted by b and h (Fig. 279) and hence the equation of the ellipse is 


b* + w =1 - 


A strip parallel to the y -axis will be selected as the element of area. From the 
equation of the ellipse, y may be expressed in terms of x by the equation, 


y = j'\/b 2 —x 2 . 
b 


Hence, 

To find x; 


a= Jy y ^ x= \^ 0 — x 2 dx = \irbh. 

x=fa 

x • ydx = -jf x's/b 2 —x 2 dx = \b 2 h. 


\tv bh'X= I xda 


Therefore, 


\b 2 h 46 


























FIRST MOMENTS AND CENTROIDS OF AREAS 395 


To find y; the same strip will be used for the element of area, its centroid 
being at the distance \y from the rr-axis. Thus, 

\irbh-y = ^ ^y'y dx= ^ 2 Jo (b 2 —x 2 )dx = %bh 2 . 

Therefore, 

-_lbh 2 4 h 
^ \irbh 37T 


Problem 249—Parabolic Segment.—Let the segment be bounded by 

h 2 x 

the z-axis, the line x = a, and the parabola y 2 = — as shown in Fig. 280. A 



strip parallel to the y-axis will be selected as the element of area, the area of 
the strip being expressed by ydx. The area of the segment, then, is 


To find x; 


Therefore, 




b 2 
x 3/ ^dx=-hb 2 . 
5 


-J hh2 J hh2 _\ 

a %bh 5 


To find y ; the same elementary strip will be selected, but since each point 
of the element is not the same distance from the rr-axis, its moment must be 
expressed as the product of the area of the strip and its centroidal distance,. 

from the rc-axis. Thus, 

2 















396 


APPENDIX I 


Therefore, 

_ \h 2 b Wb S 
V %bh~ 8 

Problem 250. Shaded Area in Fig. 281.—Let it be required to locate the 

centroid of the area between the 
y-axis, the line y = h, and the parabola 
h 2 x 

y 2 =— represented by the shaded area 
b 

in Fig. 281. 

The element of area shown in the 
figure is parallel to the rr-axis, and is 
equal to xdy. The shaded area a is 
the area (bh) of the rectangle minus 
Fig. 281. the area (f bh) of the parabola segment. 

Hence a = %bh. 

To find x: In accordance with the second of the rules in Art. 159, we have 



Therefore 


To find y: 


Therefore, 


-if 


« 4 6 2 1 b 2 h 5 1 

—dy = -=— b 2 h. 

h* y 2 h* 5 10 


- 3 1, 
* = !o 6 ' 


ay 


jrto-f 

u 


yxdy 
' b h* bh 2 

y3dV = h* 4 _ T 


y=l h - 


160. Centroids of Composite Areas.—As noted in Art. 158, 
if the centroid of an area is known, the moment with respect to 
an axis is most easily found by multiplying the area by the dis¬ 
tance of the centroid from the axis. Thus, if a given area can be 
divided into parts, the centroids of which are known, the moment 
of the whole area may be found, without integrating, by obtaining 
the algebraic sum of the moments of the parts into which the area 
is divided, the moment of each part being the product of that part 
and the distance of its centroid from the line. Thus, for example, 
if a'i, a' 2 , cl ' 3, etc., denote the parts into which an area a is divided, 










FIRST MOMENTS AND CENTROIDS OF AREAS 397 


and x'o, x"o, x'"o, etc., denote the ^-coordinates of the centroids of 
the respective parts, then, 

(a'i+a , 2+a , 3+ . . . )x=a'ix'o~\~d r 2X / 'o~{~ci r sx" / o, . (243) 


or, 

Similarly, 


ax = X(a'x o). 
ay = 2 (o'Vo). 


ILLUSTRATIVE PROBLEM 


Problem 251. Locate the centroid of the T-section shown in Fig. 282. 


Solution .—If axes be selected as indi¬ 
cated, it is evident from symmetry that 
x = 0. By dividing the given area into areas 
a'\ and a\, and by taking moments about 
the bottom edge of the area, y may be found 
as follows: 


ay = ’2 (a'yo), 

- 12X1+12X5 
V ~ 6X2+6X2 



PROBLEMS 


252. Locate the centroid of the channel section shown in Fig. 283. 



Ans. z = 0.79 in. 



253. Locate the centroid of the shaded area shown in Fig. 284. 

254. Locate the centroid of the segment of a circle as shown in Fig. 285. 





























398 


APPENDIX I 


In the expression for x make « = - an< ^ see ^ ^he resu ^ a S rees with the result 
found in Prob. 247 for a semicircle. 


255. Find the distance, from the larger base, of the centroid of the area of 

h 2&1-P&2 


the trapezoid shown in Fig. 286. 


Ans. v = -- 

3 61+62 




256. Fig. 287 represents the cross-section of the end post of a bridge. The 
area of each channel section is 4.78 square inch. Find the distance from the 
top of the cover plate to the centroid of the section. 

















APPENDIX II 


SECOND MOMENT OR MOMENT OF INERTIA OF AN AREA 

161. Moment of Inertia of an Area Defined.—In the analysis of 
many engineering problems as, for example, in determining the 

stresses in a beam or a shaft, expressions of the form Jx 2 da are 

frequently met, in which da represents an element of an area a, 
and x is the distance of the element from some axis in, or perpen¬ 
dicular to, the plane of the area, the limits of integration being such 
that each element of the area is included in the integration. An 
expression of this form is called the second moment of the area or the 
moment of inertia of the area with respect to the given axis. 

The moment of inertia of an area with respect to an axis in, or 
perpendicular to, the plane of the area may, then, be defined as the 
sum of the products obtained by multiplying each element of 
the area by the square of its distance from the given axis. 

The term moment of inertia is somewhat misleading, since 
inertia is a property of physical bodies, only, and hence an area 
does not possess inertia. For this reason the term, second moment 
of an area, is to be preferred, particularly when contrasting the 
expressions of the form here discussed with expressions which were 
defined as first moments of areas in Appendix I. It may be noted 
that each term x 2 da in the summation can be written in the form 
x(xda), and hence represents the moment of the moment of an 
element of area, that is, the second moment of the element. The 
term moment of inertia, however, is very widely used, due to the 
fact that the expression is of the same form as an expression which 
is defined as the moment of inertia of a body and which does have 
a physical significance. 

The moment of inertia of an area with respect to an axis will 
be denoted by I for an axis in the plane of the area and by J 
for an axis perpendicular to the plane of the area. The particular 

399 



400 


APPENDIX II 


axis (or direction of the axis) about which the moment of inertia 
is taken will be denoted by subscripts. Thus, the moments of 
inertia of the area (Fig. 288) with respect to the x- and y -axes 
y are expressed as follows: 


I x =jy 2 da, 
and 

I v = jx 2 da. 

Fig. 288. 


(244) 



Units and Sign .—Since the moment of inertia of an area is the 
sum of a number of terms each of which is the product of an area 
and the square of a distance, the moment of inertia of an area 
is expressed as a length to the fourth power. If, then, the inch 
(or foot) be taken as the unit of length, the moment of inertia 
will be expressed as inches (or feet) to the fourth power (written 
in. 4 or ft. 4 ). Further, the sign of each of the products x 2 da is 
always positive since x 2 is always positive, whether x is positive 
or negative, and da is essentially positive. Therefore the moment 
of inertia, or second moment, of an area is always positive. In 
this respect it differs from the first moment of an area, which may 
be positive, negative, or zero, depending on the position of the 
moment axis. 

162. Polar Moment of Inertia.—The moment of inertia of an 
area with respect to a line perpendicular to the plane of the area is 
called the polar moment of inertia of the area and, as noted in Art. 
161, will be denoted by J. Thus the polar moment of inertia 
with respect to the 2 -axis, of an area in the #?/-plane (Fig. 288) 



Jz = I v +I x .(245) 

Hence the following proposition may be stated: 













MOMENT OF INERTIA OF AN AREA 


401 


The polar moment of inertia of an area with respect to any axis 
is equal to the sum of the moments of inertia of the area with respect 
to any two rectangular axes in the plane of the area which intersect 
on the given polar axis. 

163. Radius of Gyration.—Since the moment of inertia of an 

area x 2 da or J' r 2 da, etc.,^ is four dimensions of length, it 

may be expressed as the product of the total area, a, and the 
square of a distance, k. Thus, 


or 



J 2 = J'r 2 da = ak z 2 . 


(246) 


The distance k is called the radius of gyration of the area with 
respect to the given axis, the subscript denoting the axis with 
respect to which the moment of inertia is taken. The radius of 
gyration of an area with respect to a line, then, may be defined as 
a distance such that, if the area were conceived to be concen¬ 
trated at this distance from the given line, the moment of inertia 
would be the same as the moment of inertia of the actual or dis¬ 
tributed area with respect to the same line. 

From the equation I v =j'x 2 da = ak y 2 , it will be noted that 

k v 2 , the square of the radius of gyration with respect to the y- axis, 
is the mean of the squares of the distances, from the ?/-axis, of the 
equal elements of area into which the given area may be divided, 
and that it is not the square of the mean of these distances. The 
mean distance (x) of the elements of area from the y- axis is the 
centroidal distance as discussed in the preceding chapter. Hence 
ax 2 does not represent the moment of inertia of an area with respect 
to the y- axis. 

164. Parallel Axis Theorem.—If the moment of inertia of an 
area with respect to a centroidal axis in the plane of the area is 
known, the moment of inertia with respect to any parallel axis in 
the plane may be determined, without integrating, by means of a 
proposition which may be established as follows: In Fig. 290 let 
YY be any axis through the centroid, C, of an area and let Y'Y' be 
any axis parallel to YY and at a distance d therefrom. Further, 



402 


APPENDIX II 


let the moment of inertia of the area with respect to the axis YY 
be denoted by 7 and the moment of inertia with respect to Y'Y' by 
7. By definition then, 

I = f(x+d) 2 da 


Therefore, 


= j~ x 2 da +2 djxda + d 2 J da. 

I=I-\-ad 2 , since Jxda = ax = C 


• • (247) 


Hence the following proposition may be stated: 

The moment of inertia of an 
area with respect to any axis in the 
plane of the area is equal to the 
moment of inertia of the area with 
respect to a parallel centroidal axis 
plus the product of the area and 
the square of the distance between 
the two axes. This proposition is 
called the parallel axis theorem. 

A corresponding relation exists between the radii of gyration of 
the area with respect to two parallel axes, one of which passes 
through the centroid of the area. For, by replacing I by ak 2 and 
7 by ak 2 the above equation becomes, 



ak 2 = ak 2 -\-ad 2 . 

Whence, 

k 2 =~k 2 +d 2 , .(248) 


where k denotes the radius of gyration of the area with respect to 
any axis in the plane of the area and k denotes the radius of gyra¬ 
tion of the area with respect to a parallel centroidal axis. 

Similarly, for polar moments of inertia and radii of gyration, it 
can be shown that, 

J—J -\-ad 2 , 

and, 


k — k 2 -\~d 2 } 


where J and k denote the polar moment of inertia and radius of 
gyration, respectively, of the area with respect to the centroidal 
axis and J and k denote the polar moment of inertia and radius of 









MOMENT OF INERTIA OF AN AREA 


403 


gyration, respectively, of the area with respect to an axis parallel 
to the centroidal axis and at a distance d therefrom. 

165. Moments of Inertia Found by Integration.—In deter¬ 
mining the moment of inertia of a plane area with respect to a line, 
it is possible to select the element of area in various ways and to 
express the area of the element in terms of either Cartesian or polar 
coordinates. Further, the integral may be either a single or double 
integral, depending on the way in which the element of area is 
selected; the limits of integration are determined, of course, 
from the boundary curve of the area. In any case, however, 
the elementary area must be taken so that: 

(1) All points in the element are equally distant from the axis 
with respect to which the moment of inertia is to be found, other¬ 
wise the distance x in the expression x 2 da would be indefinite. 
Or, so that, 

(2) The moment of inertia of the element, with respect to the 
axis about which the moment of inertia of the whole area is to be 
found, is known, the moment of inertia of the area then being 
found by summing up the moments of inertia of the elements. 
Or, so that, 

(3) The centroid of the element is known and also the moment 
of inertia of the element with respect to an axis which passes 
through the centroid of the element and is parallel to the given 
axis; the moment of inertia of the element may then be expressed 
by means of the parallel axis theorem. 

The moments of inertia of some of the simple areas will now be 
found in the following illustrative problems: 

ILLUSTRATIVE PROBLEMS 

Problem 257.—Determine the moment of inertia of a rectangle, in terms 
of its base b and altitude h, with respect to (a) a centroidal axis parallel to 
the base; (6) an axis coinciding with the base. 

Solution .—(a) Centroidal Axis.—The element of area will be selected in 
accordance with rule (1) above, as indicated in Fig. 291. The moment of 
inertia of the rectangular area with respect to the centroidal axis, then, is, 

+- 

7 x=z f y2da= J h y2bdy 

~~2 



404 


APPENDIX II 


(6) Axis Coinciding with the Base. First Method .—The element of area 
will be selected as indicated in Fig. 292. The moment of inertia of the rect¬ 
angle with respect to the base, then, is, 


h=j' y2da = J Q 


= ! y*bdy 


M*. 




[Av 


/ 

? 

f da=bdy 

Si 

< 

V 

<- b -> 



Fig. 292. 


Second Method .—Since the moment of inertia of the rectangle with respect 
to a centroidal 'axis is Tzbh 3 , the moment of inertia with respect to the base 
may be found from the parallel axis theorem (Art. 164). Thus, 


lb = I x~\~n 


/A 2 


=&bv+bhx~ 

=lbhK 

Problem 258.—Determine the moment of inertia of a triangle, in terms of 
its base b and altitude h, with respect to (a) an axis coinciding with its base; 
(6) a centroidal axis parallel to the base. 

Solution. —(a) Axis Coinciding with the Base.—The elementary area 
will be selected as shown in Fig. 293. The moment of inertia of the area of the 
triangle with respect to the base, then, is, 


h = j”y 2 da = j~y*xdy. 


But, from similar triangles, 


x h-y 
b~ h ’ 


x = T (h-y). 
h 


Hence, 





















MOMENT OF INERTIA OF AN AREA 


405 


Therefore, 





k- b -*f 


b -H 


Fig. 293. 


Fig. 294. 


(b) Centroidal Axis Parallel to the Base.—-The centroidal axis parallel 
to the base (axis XX) is shown in Fig. 294 (see Prob. 246). Using the parallel 
axis theorem, the moment of inertia of the triangular area with respect to the 
centroidal axis is, 


Ix=h~a{\hY 
= t hbh*-hbhXW 



Problem 269. Determine the moment of inertia of the area of a circle, in 
terms of its radius r, with respect to an axis coinciding with the diameter; 
(a) using Cartesian coordinates; (6) using polar coordinates. 

Solution. —(a) Cartesian Coordinates.—The element of area will be 
selected as shown in Fig. 295. The moment of inertia of the circular area 
with respect to the diameter, then, is, Y 




Fig. 295, 


( b ) Polar Coordinates.— The element of area will be selected as shown in 
Fig. 296. Hence, 



















406 


APPENDIX II 



-XX 
-XX 

r 4 C 2x • 

= I sin Odd 

4 Jo 


(p sin 0) 2 pdpdO 


p 3 sin 2 0 dpdO 


r 4 11 
= — Xir = -7rr 4 = — ird 4 . 

4 4 64 


Problem 280.—Determine the polar moment of inertia of the area of a 
circle of radius r with respect to a centroidal axis: (a) by integration; (6) 
by use of the theorem of Art. 132. 

Solution. —(a) By Integration.—By selecting the element of area as 
indicated in. Fig. 297, the polar moment of inertia of the circular area is, 



(b) By Use of Theorem of Art. 162.—Since I x and I y are each equal to 
\-irr 4 (Prob. 259), the polar moment of inertia of the area of the circle is, 

J z = Ix~\~Iy 

= £7rr 4 + £7rr 4 



PROBLEMS 

261. Determine the moment of inertia of the area of a circle, with respec, 
t,o an axis tangent to the circle, in terms of, r, the radius of the circle. 

262. Determine the polar moment of inertia of the area of a rectangle of 
base b and altitude h with respect to the centroidal axis. 

Ans. J =- 1 \6/i(5 2 +/i 2 ). 

263. Find the moment of inertia and radius of gyration of a circular area 
16 in. in diameter, with respect to a diameter. 

264. Determine the moments of inertia of the area of an ellipse, the prin¬ 
cipal axes of which are 2 b and 2 h, with respect to the principal axes. 

Ans. Ij) = \Trbh 3 . h = \M 3 . 




MOMENT OF INERTIA OF AN AREA 


407 


265. The base of a triangle is 8 in. and its altitude is 10 in. Find the 
moment of inertia and radius of gyration of the area of the triangle with 
respect to the base. 

266. Find the polar moment of inertia and radius of gyration of the area 
of a square, each side of which is 15 in., with respect to an axis through one 
corner of the square. 

267. Find the polar moment of inertia, with respect to a centroidal axis, 
of the area of an isosceles triangle having a base b and altitude h. 

Ans. J =Tzbh(lb 2 -\-$h 2 ). 

166. Moments of Inertia of Composite Areas. —When a com¬ 
posite area can be divided into a number of simple areas, such as tri¬ 
angles, rectangles, and circles, for which the moments of inertia 
are known, the moment of inertia of the entire area may be 
obtained by taking the sum of the moments of inertia of the several 
areas. Likewise, the moment of inertia of the part of an area 
that remains after one or more simple areas are removed may be 
found by subtracting, from the moment of inertia of the given area, 
the sum of the moments of inertia of the several parts removed. 


ILLUSTRATIVE PROBLEMS 


Problem 268. Locate the horizontal centroidal axis, XX, of the T-section 
shown in Fig. 298 and find the moment of inertia of the area with respect to 
this centroidal axis. 


Solution. First Method .—The distance, y, of the centroid of the area from 
the axis X iX x may be found from the equation, 


Thus, 


ay = X(a'y 0 ). 

__ 12X7 + 12X3 
V ~ 12 + 12 


= 5 in. 


The moment of inertia with respect to the XX 
axis is the sum of the moments of inertia of the 
three parts a'i, a' 2 , and a' 3 , with respect to that 
axis. Thus, 



7 a: = iVX6X(2) 3 +12X(2) 2 +^X2X(l) 3 +|X2X(5) 3 


= 4+48+.67 +83.33 
= 136 in. 4 


















408 


APPENDIX II 


Second Method .—The moment of inertia of the T-section may also be 
determined as follows: First find the moment of inertia of the T-section with 
respect to the axis X 1 X 1 by subtracting the moments of inertia of the parts 
a\ and a '5 from the moment of inertia of the rectangular area ABCD, and then 
find I x for the T-section by use of the parallel axis theorem. Thus, the 
moment of inertia, I x , of the T-section with respect to the XiX 1 axis is, 


/x = £X6X(8) 3 —2X£X2X(6) 3 = 736 in. 4 ; 

and 

I x =I x —Ad 2 = 736 - 24 X (5) 2 = 136 in. 4 . 


Problem 269. Find the moment of inertia of the channel section shown in 
Fig. 299 with respect to the line XX. Find also the moment of inertia with 
respect to the parallel centroidal axis. 



Solution .—The area may be divided into triangles and rectangles as shown 
in the figure. The values used in the solution may be put in tabular form as 
shown below, where a' denotes the area of any part, y 0 the distance of the cen¬ 
troid of the part from the line XX, I 0 the moment of inertia of the part with 
respect to its own centroidal axis parallel to XX, and I' x the moment of inertia 
of the part with respect to the axis XX. 


Part 

a' 

Vo 

a'yo 

h 

a'yo 2 

I' x = Io+ayo 2 

a'i 

0.745 

1.61 

1.20 

0.44 

1.93 

2.37 

a\ 

.745 

1.61 

1.20 

• .44 

1.93 

2.37 

a ' 3 

.585 

1.17 

0.68 

.23 

0.80 

1.03 

a\ 

.585 

1.17 

.68 

.23 

.80 

1.03 

a's 

.360 

0.14 

.47 

.02 

.07 

0.09 


6.02 in. 2 


4.23 in. 3 



6.89 in. 4 


Thus the moment of inertia I x of the area with respect to the XX axis is, 


I x = Xl' x = 6.89 in. 4 


























MOMENT OF INERTIA OF AN AREA 


409 


Further, the total area is a = 2a' =6.02 in. 2 and the moment of the area with 
respect to the XX axis is 2(a'?/o) =4.23 in. 3 Hence, the distance, y, of the 
centroid of the area from the XX axis is, 




Therefore, the moment of inertia with respect to a line through the centroid 
and parallel to XX is given by the equation, 

lx = lx — 


= 6.89—6.02 X(.70) 2 


= 6.89-2.95 


= 3.94 in. 4 



410 


APPENDIX II 


PROBLEMS 

270. A wooden column is built up of four 2-in. by 8-in. planks as shown 
in Fig. 300. Find the moment of inertia of the cross-section with respect to 
the centroidal axis XX. Ans. 7 a; = 981 in. 4 



K- 8 - - ->\ 

Fig. 300. 



271. Find the moment of inertia of the angle section (Fig. 301) with respect 
to each of the centroidal axes parallel to the two legs of the angle. 


272. In Fig. 302 is shown the cross-section of a standard 9-in. 21-lb. I- 
beam (fillets are neglected). Find the moments of inertia of the section with 
respect to the centroidal axes, XX and YY. Ans. 7^ = 84.9 in. 4 ; 7^ = 5.16 in. 4 



Fig. 302. 



273. In Fig. 303 is shown the cross-section of a standard 3| in. by 5-in. 
Z-bar (fillets are neglected). Find the moments of inertia of the section with 
respect to the centroidal axes XX and YY. 

274. In Fig. 304 is shown a built-up section made of a ^-in. by 20-in. plate 

and four angles. Find the moment of inertia of the section with respect to 
the XX axis. Ans. 7a; = 1850 in. 4 




































MOMENT OF INERTIA OF AN AREA 


411 



!Y 



275 . In Fig. 305 is represented a 16-in. circular plate in which there are 
drilled five 2-in. holes and one 4-in. hole as shown. Find the moment of inertia 
of the area of the holes with respect to the XX axis and also with respect 
to the YY axis. Ans. I x = 252 in. 4 

167. Approximate Method.—It is sometimes necessary to 
determine the moment of inertia of an area that has a boundary 
curve which cannot be defined by a mathematical equation. An 
approximate value of the moment of inertia of such an area may 
be found by the following method. For convenience, however, a 
simple area will be selected so that the approximate value of the 
moment of inertia as determined by this method may be com¬ 
pared with the exact value. Thus, 
let the moment of inertia of the area 

of a rectangle, with respect to an axis 1 

coinciding with its base, be found. : 

The area may be divided into any con¬ 
venient number of equal narrow strips Fig. 306. 

parallel to the base, as shown in Fig. 

306. (The narrower the strips the more closely will the result 
agree with the exact result.) Let the area be divided into ten 
such strips each 0.2 in. in width. The moment of inertia of the 
rectangle is equal to the sum of the moments of inertia of the strips. 
The moment of inertia of any particular strip with respect to the 
base of the rectangle is 

•A-X6X(!) 3 +6 XiXy 2 , 

where y is the distance of the centroid of the particular strip from 
the base. The first term is small and may be omitted without 































412 


APPENDIX II 


serious error. The moment of inertia of each strip then is approx¬ 
imately equal to the product of the area of the strip and the 
distance of its centroid from the base. Hence, the moment of 
inertia of the rectangle is, 

7 = f(.l 2 +.3 2 +.5 2 +.7 2 +.9 2 +l.l 2 +1.3 2 +1.5 2 +1.7 2 +1.9 2 ) 

=f X13.3 
= 15.96 in. 4 . 

According to Prob. 257 the exact value is, 


1 = %bh 3 = JX6X2 3 = 16 in. 4 


APPENDIX III 


LOCALIZED STRESS; ITS OCCURRENCE, SIGNIFICANCE 
AND MEASUREMENT 

By H. F. Moore 1 and F. B. Seely 


168. Limitations of the Ordinary Methods of Analysis in 
Mechanics of Materials.—In Art. 126 and 143 it was pointed out 
that the use of the analyses and formulas given in the mechanics 
of materials as developed in Part I does not make possible the com¬ 
putation of all stresses existing in a machine part or a structural 
member. However, the ordinary formulas of the mechanics of 
materials are not to be thought of as invalid, but as presenting an 
incomplete statement or picture of the stress conditions in a mem¬ 
ber. For example, in an I-beam subjected to bending, neither 
/ M c \ 

the flexure formula ( s=—^,Art. 341, which gives the tensile and 
compressive stresses, nor the formula for shearing stress ( s 8 = 
Jijzda, Art. 40^ take account of the intensity of bearing stress at 
the edge of a bearing block; in a shaft subjected to torsion the torsion 
formula ( s » = ~j} Art- 26 J cannot be used to determine the stress at 
the root of a key way; in a bolt subjected to an axial tensile load, 


the axial load formula = —, Art. 3 J does not give a measure of the 

concentration of stress at the root of the thread, even when a 
is taken as the area at the root of the thread, etc. 

169. Definition and Illustrations of Localized Stress.—Local¬ 
ized stress may be defined as stress existing over a small area of a 
machine part or a structural member, which stress is not deter- 


1 Research Professor of Engineering Materials at University of Illinois. 

413 



414 


APPENDIX III 


mined by the ordinary formulas of mechanics of materials as devel¬ 
oped in Part I. Localized stresses occur' at (1) the fillet of a 
shoulder in a shaft, (2) at the edge of a rivet hole, (3) at the root of 
the thread of a bolt, (4) at the edge of a non-metallic inclusion in a 
piece of steel, (5) at the edge of a bearing block, (6) at the bearing 
area between a car wheel and the rail, (7) at the root of a deep 
tool mark in a shaft, (8) at the sharp corner at the junction of head 
and shank of a bolt, etc. 

170. The Significance of Localized Stress.— (a) In Members 
Subjected to Static Loading. —As was discussed in Art. 126 and 143, 
for machine parts and structural members subjected to steady 
load, or to load repeated a few times, localized stresses are, in gen¬ 
eral, of very little significance, unless the members are made of 
brittle material; ductile material yields at the points of high 
localized stress and hence the stress is transferred from the over¬ 
stressed fibers to adjacent understressed fibers, and tho deforma¬ 
tion of the member as a whole is not appreciably affected. There¬ 
fore, when members made of ductile material are subjected to 
static loads the ordinary formulas of mechanics of materials may 
be regarded as giving a complete picture of the significant stresses 
in the more simple structural and machine members, although the 
secondary and localized stresses may be of much importance 
in the more complex structures, particularly those of large size 
and of new type. 

( b ) In Members Subjected to Impact Loading. —In machine parts 
or structural members subjected to impact or energy loading, 
localized stress is of more importance than it is in members sub¬ 
jected to steady loads. As discussed in Chapter XIII, the energy 
absorbed by a material when it is stressed is proportional to the 
square of the stress. This means that the small portions of a member 
where the localized stresses occur absorb an excessive amount of 
the energy load before the main portion of the member can be 
appreciably stressed and hence before the main portion can be 
made to absorb an appreciable share of the energy load. As a 
result, the small portion where a localized stress occurs is likely 
to be stressed above the yield-point of the material; this in turn 
still further localizes the absorption of energy, and there is 
danger of rupture even if the material is relatively ductile. The 
stresses developed by energy loads are discussed in Chapter 
XIII. 


LOCALIZED STRESS 


415 


(c) In Members Subjected to Repeated Loading .—As discussed 
in Chapter XIV in machine parts and structural members that 
are subjected to repeated loading, localized stresses must fre¬ 
quently be considered to be the significant stresses, and hence the 
ordinary formulas in mechanics of material give only a very rough 
estimate of the values of the stresses that may cause damage to 
the material. 

The above discussions indicate the need for methods of deter¬ 
mining the magnitudes of localized stresses; some of these methods 
are discussed below. 

171. Methods of Determining Magnitude of Localized Stress.— 

(a) Mathematical Analysis. —The simple mathematical analyses 
given in Part I serve to determine the average stresses over small 
areas of many common structural members and machine parts, but 
they do not determine the maximum stress at a point. When by the 
use of mathematical analyses it is attempted to get a more com¬ 
plete picture of stress distribution than is given by the ordinary 
analysis of mechanics of materials, especially when it is attempted 
to determine stresses or strains at sudden changes of outline of a 
part,—at fillets, shoulders, threads, and holes,—then the differ¬ 
ential equations involved in the mathematical analysis become 
very difficult, if not impossible, of solution. 

As an illustration of the determination of localized stress by 
means of mathematical analysis the reader is referred to the study 
by Inglis of the stresses at the edge of cracks in a plate. 1 In this 
paper mathematical analysis is applied to the determination of 
stresses and strains at the edges of elliptical holes in a plate, and a 
crack is regarded as an elliptical hole with a very short minor axis. 

At the present development of our knowledge of mathematics, 
mathematical analysis can not be regarded as an available tool for 
the complete determination of localized stress in many machine 
parts and structural members. 

(6) Tests to Destruction of Models Made of Brittle Material .— 
A method of stress analysis, which has been used with good suc¬ 
cess to give approximate values of the maximum unit-stress in 
complex shapes, consists in loading a model of the member made 
of a brittle material that has a flat (nearly straight) stress-strain 

1 Inglis, C. E. “ Stresses in a plate due to the Presence of Cracks and 
Sharp Corners,” Trans. Inst, of Naval Architects (British), Vol. LV, Pt. I, p. 
219 , 1913 . 


416 


APPENDIX III 


diagram up to the ultimate (such a material as hard cast iron, 
plaster of paris, or glass); from the same material is also made a 
model of simple shape, most commonly a tension test specimen. 
The specimen of simple shape and the model of complex shape are 
then tested to destruction; the test of the simple shape gives the 
ultimate strength of the material, and the test of the model gives 
with a fair degree of accuracy the load which produces this ulti¬ 
mate stress in the most stressed fibers of the model, and from these 
results the relation between load and maximum unit stress can 
be computed. The use of this method assumes that the deviation 
of the stress-strain diagram of the material from a straight line 
may be neglected. 

An example of this method of stress-determination is found in 
the determination, first by Bach 2 and later by Kommers 3 of the 
shearing stresses in torsional members having non-circular cross- 
sections. Both Bach and Kommers used cast-iron specimens. 

This method must be regarded as yielding only approximate 
results. The deviation from Hooke’s law is considerable for all 
known materials when stressed up to rupture, and the effect of 
lateral restraint at the points of maximum (localized) stress causes 
the apparent values of the maximum unit-stress found in the test 
to be less than the true values. Moreover, by this method only 
the maximum unit-stress can be determined; this maximum unit- 
stress is, however, usually the most important stress to be deter¬ 
mined. This method is simple, the model specimens are not very 
expensive to prepare, and valuable results may be secured by its 
use. 

(c) Yield-point Tests of Models Made of Ductile Material .— 
The maximum localized stress in a member may be determined by a 
method somewhat similar to ( b ), in which a model of the piece made 
of ductile material with a well-marked yield-point is employed, 
such as ordinary low-carbon steel. The'yield-point of the material 
is determined by the test of a tension specimen, a compression 
specimen, or a hollow torsion specimen depending on the kind of 
stress to which the model is subjected, and then load is applied to 
the model of the desired shape made of the same material. As load 
is applied the surface of the model is carefully watched for evidence 

2 Bach, C. “ Elastizitat und Festigkeit,” Eighth edition, pp. 342 to 401. 

3 Kommers, J. B. “Torsion Tests of Cast Iron,” American Machinist, 
May 28, 1914, p. 941. 


LOCALIZED STRESS 


417 


of yielding. If the model is flat and if a surface can be left with 
the mill scale on it, the flaking off of this mill scale gives a fairly 
accurate indication of a localized stress equal to the yield-point of 
the material, and from the known value of the yield-point of the 
material and of the load on the model when flaking first occurs, the 
relation between load and unit-stress can be determined. If the 
surface of the model is painted with a wash of white Portland 
cement the yielding can be detected by the cracking of the cement. 

Instead of using the flaking of mill scale or of cement paint, 
as an indicator of the yield-point stress, the surface of the model 
may be polished, and the localized yield-point stress detected by a 
slight wrinkling of the polished surface, giving rise to what are 
known as “ Luders’ lines.” 

The yield-point method 4 of determining localized stress, like 
the brittle material method (6), determines only the maximum 
unit stress. It is usually desirable to use rather large models, 
and this necessitates the use of a large testing machine to develop 
in them the yield-point stress. This method, like method (6), 
should be regarded as yielding only approximate results, but it is 
very useful when more precise methods are not feasible. 

(d) Tests of Models Made of Plastic Material .—Certain soft 
metals, such as soft copper, lead and type metal, acquire a per¬ 
manent set when subjected to a very small stress, and the perma¬ 
nent set is approximately proportional to the stress applied. An 
approximate determination of the localized stress in irregular¬ 
shaped members can be made by making models of the irregular 
members out of soft copper, type metal or other plastic metal, 
laying off on the surface of the model a series of reference lines at 
known distances from each other, applying load, and, after remov¬ 
ing the load, measuring the distortion between reference lines at 
various parts of the member. This measurement can sometimes 
be best made with the aid of a low-power microscope. 

The relative distortion between gage lines at various parts of 
the model gives an approximate measure of the proportionate 
unit-strain developed at that location. Shearing unit-strain is 
measured by angular distortion between reference lines, tensile 
unit-strain or compressive unit-strain is measured by change of 

4 This method was used to determine the maximum stress in eye-bars 
by Epstein and Schwartz; the results of their tests were presented in 1911 as a, 
student’s thesis at the University of Illinois. 


418 


APPENDIX III 


distance between lines. If the strain-distribution is known the 
stress-distribution can also be determined since the relation of 
stress to strain is known. At boundaries of the model this deter¬ 
mination of stress from strain is usually quite simply accom¬ 
plished. 

In a distorted plastic model the distortions to be measurable 
must be of appreciable magnitude, and the distortions will them¬ 
selves somewhat modify the distribution of stress throughout the 
model. This method of stress-analysis gives some idea of the 
general distribution of stress over the entire surface of the member, 
as well as the location and approximate determination of the max¬ 
imum unit-stress. In using this method it is desirable to use as 
large models as is feasible, so that the distortion becomes measur¬ 
able over a short gage length between reference lines. 

The use of the distortion of plastic models in determining stress- 
distribution is illustrated in Bach’s treatment of non-circular mem¬ 
bers subjected to torsion. 5 

(e) Tests with Rubber Models .—A method somewhat similar to 
id) employs a rubber model similar in shape to the member to be 
studied. On the surface of the model reference lines are laid off, 
and the distortion between reference lines is measured while the 
member is resisting the load. This method is subject to the same 
inaccuracies as is method (d), and in addition is subject to error 
due to the fact that the stress-strain diagram for rubber deviates 
appreciably from a straight line, even for low stresses. The method 
is, however, very well adapted to show in a striking manner the 
general scheme of stress-distribution over the surface of a member. 
In using this method it is desirable to use as large models as is 
feasible, and, except for very simple shapes, such models are expen¬ 
sive. 

Examples of the use of rubber models in determining stress- 
distribution are furnished by the work of Chiles and Kelley 6 
on eyebars and plates with holes, and of Trelease 7 on stresses in 
flat slab structures. 

5 Bach, C. Elastizitat und Festigkeit, Eighth edition, pp. 342-350. 

6 Chiles and Kelley, “ The Resistance of Materials, The Effect of Sudden 
or Abrupt Changes in the Section on the Distribution of the Unit Stresses/' 
Railway Mechanical Engineer, March, April, May, 1919. 

7 Trelease, “The Design of Concrete Flat Slabs,” Proc. American Con¬ 
crete Institute, Vol. VIII, 1912, p. 218. 


LOCALIZED STRESS 


419 


(/) The Use of the Strain Gage on Actual Structures or on Models. 
—In any metal member the elastic strain in a gage length of 2 inches 
or more can be measured with a good degree of accuracy by means 
of a strain gage. 8 The strain gage is a special form of micrometer 
for measuring changes of length along a gage line on the surface 
of a member or specimen. The use of the strain gage to determine 
localized stress is especially applicable to full-size structural mem¬ 
bers, and large parts of machines. The instrument has been used 
successfully in determining strains in reinforced concrete structures, 
steel bridge members, car bolsters, built-up girders, and steel 
columns. 

A limitation of the strain-gage method is the relatively long 
gage length which must be used. The ordinary strain gage cannot 
be used to determine localized stress at fillets, at the root of threads 
or at grooves in a shaft. The train-gage method is especially 
promising in the field of structural engineering. 

An illustration of the use of the strain gage in determining 
localized stress is to be found in the work of Moore and Wilson on 
stresses in the webs of I-beams and girders. 9 

By the use of special very delicate extensometers attached to a 
specimen it is possible to measure strains over much shorter gage 
lengths than 2 inches. The investigations by Preuss of stress 
distribution round notches and holes in flat bars furnish a striking 
illustration of this method of stress analysis. 10 

(g) The Use of Transparent Models , and Polarized Light .— 
An elegant and accurate method for the determination of stress 
distribution in flat members subjected to stress in one plane 
employs model specimens made of glass, celluloid, or other trans¬ 
parent material, which are viewed by polarized light. Space for¬ 
bids any detailed discussion of the optical problems involved in 

8 For a description of the strain gage and a discussion of the technique 
of its use, see Slater and Moore, “ The Use of the Strain Gage in Testing 
Materials,” Proc. Am. Soc. for Testing Materials, Vol. XIII, p. 1019 
(1913). 

9 Moore and Wilson, “ The Web Strength of I-beams and Girders,” 
Bulletin 86, Engineering Experiment Station, University of Illinois. 

10 Preuss, E. “ Versuche liber die Spannungsverteilung in Gelochten 
Zugstaben,” Zeit. Ver. Deut. Ing., 1912, p. 1780. 

“ Versuche liber die Spannungsverteilung in Gekerbten Zugstaben,” 
Zeit. Ver. Deut. Ing., 1913, p. 664. See also Johnson’s Materials of Construc¬ 
tion, fifth edition, p. 663. 


420 


APPENDIX III 


the method. 11 In general, if a transparent substance is viewed 
by polarized light it appears to be of some definite color, and the 
color depends on the state of strain in the material. If the stress- 
distribution across a section is uniform the color will be uniform; 
if the stress distribution is variable there will be bands of various 
colors, merging into each other; if there are sudden changes 
of stress these bands are so close together that sharply marked 
dark spots show up. A glance at a stressed model specimen 
illuminated by polarized light shows whether there are points of 
high localized stress. A simple method of estimating the magni¬ 
tude of localized stress consists in making from the same material 
as the model specimen a tension (or compression) test of a specimen 
of simple shape, in which the uniformity of color across the sec¬ 
tion shows a uniform stress-distribution. This auxiliary specimen 
is loaded until it shows some definite color red, say; it is then loaded 
further until the color changes through the spectrum to red again, 
and the difference in load for the change gives a measure of the 
stress required in the material to cause a “ red-to-red ” change. 
The model specimen of the shape to be studied is then tested, 
and the load necessary to cause a “ red-to-red ” change of color 
at any desired location is determined. This load then produces at 
the particular location a stress of known magnitude, and from the 
relation between the loads the localized stress at that location can 
be determined. Strictly speaking, it is not the stress which is 
determined directly, but the difference between principal strains 
at the location studied. However, at the boundaries, where the 
maximum localized stress frequently occurs, one principal strain is 
zero in which case the stress is found directly. In determining 
the maximum stress at a point not on the boundary it is neces¬ 
sary to measure the change in thickness at the location studied. 
This change in thickness is proportional to the sum of the prin¬ 
cipal strains at the location. Now having both the sum and the 
difference of the principal strains the maximum strain and the 
minimum strain can be determined. 

11 Coker, E. G. “ Photo-elasticity,” Engineering (London;, Jan. 6, 1911, 

p. 1. 

“ The Optical Determination of Stress,” Phil. Mag., Oct., 1910. 

Hey mans, Paul. “ Photo-elasticity and its Applications to Engineering 
Problems.” Publications of the Mass. Inst, of Tech., Serial No. 1, May, 1922. 

“ Stress Distribution in Rotating Gear Pinions as Determined by Photo¬ 
elastic Method,” Jour. Mechanical Engineering, Mar., 1924, p. 129. 


LOCALIZED STRESS 


421 


This is a very brief outline of only one of the methods used 
in the photo-elastic determination of stress,—one of the sim¬ 
plest and less accurate methods. More refined methods use 
mono-chromatic light for illuminating the specimen, and instead 
of causing a change of color in the specimen the load is applied 
until the desired location on the specimen is black, and under 
further load the location is again brought to “ black ” by a definite 
adjustment of the optical system, this adjustment being calibrated 
in terms of stress by a test on a simple tension or compression 
specimen. 

The polarized-light method of stress-analysis can be made to 
yield results of a high degree of precision. The model specimens 
may be of small size, and comparatively inexpensive. In its 
present stage of development the method is limited to the study 
of flat members subjected to stresses in the plane of their flat 
surface;—the method could not be used, for example, to deter¬ 
mine the localized stresses at the filleted shoulders of an axle. 

(h) Repeated Stress Tests .—This method has been discussed in 
Art. 130 and 131. 

(i) Combination Methods of Stress-analysis .—It is frequently 
possible to use two or more of the methods outlined in the fore¬ 
going paragraphs, as checks on each- other; sometimes certain 
constants can be determined by an experimental method, after 
which mathematical analysis becomes possible. In determining 
the stress-distribution in the head of an eyebar the location of the 
maximum stress was determined by the yield-point test (method 
(c)), and with this location determined it was possible to apply 
mathematical analysis to determine the general stress-distribution. 

A most striking illustration of the use of a combination of 
experiment and mathematical analysis is furnished by the work of 
Griffith and Taylor in studying the torsional stresses in airplane 
propeller blades. 12 The mathematical analysis of torsional stresses 
in members of irregular cross-section (such as airplane propeller 
blades) involves the use of differential equations that have not as 
yet been solved. However, it was observed that the differential 
equations for the deformations and stresses in a twisted bar of a 
given cross-section were, except for a constant term, the same 
as the differential equations for deflections and slopes at various 

12 Griffith and Taylor, Proc. Brit. Inst, of Mech. Engrs. 1917, Oct.-Dec., 
also, “ Engineering ” (London), Vol. 124, No. 3234, Dec. 21, 1917, p. 546. 


422 


APPENDIX III 


points in an elastic film stretched over a hole of the same shape as 
the cross-section of the twisted bar, and deflected by a uniformly 
distributed normal pressure. The investigators used a soap film 
as an elastic film, stretched it over a brass plate, which had in it 
an opening of the shape of the cross-section of the airplane pro¬ 
peller blade, deflected the film by exhausting the air on one side 
of the plate, measured the deflection by means of a micrometer 
fitted with a soaped needle point, measured the slopes of the 
deflected film by means of a ray of light reflected from the film 
itself, and were thus able to determine the necessary constants for 
the computation of the shearing stresses in the airplane propeller 
blade when subjected to a given twisting moment. Their paper * 
is recommended for careful study as a brilliant example of the 
combination of methods leading to a highly accurate stress anal¬ 
ysis. 

In conclusion, it is to be noted that this appendix is scarcely 
more than a list of possible methods of stress-analysis, and that 
further study, especially of the references given, will be necessary 
before the suggestions given here can become useful in actual 
determinations of localized stress. 






















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